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@ This paper contains 40 questions each, from Botany, Zoology, Physics andChemistry.

@ Each of the 160 questions carries one mark. No negative marking for wronganswers.

@ Maximum trying allowed is 3 hours (180 minuites)

@ Use Ball - Point Pen while entering the Hall - Ticket Number and filling in Part -A of the First page.

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@ Over - writing or blackening of more than one circle will not count for marks.

@ If you wish to change your answer, erase the already darkened circle completelyand then darken the appropriate circle.

@ Candidates are prohibited from carrying any sheet of paper to the ExaminationHall except the Hall - Ticket.

@ Do any rough / scratch work on the Test Paper itself.

@ Calculators, watches with calculators, pagers & cellular phones will not be allowedinto the Examination Hall.

@ Candidates have to write suitable answers on the Answer Sheet only.

@ Candidates have to return the Answer Sheets and the Question Papers at thetime of leaving the Examination Hall.

@ Candidates can leave the Examination Hall only in the last half - an hourbefore the close of the Test.

@ Total number of pages of the Test is 30.

INSTRUCTIONS TO CANDIDATES

EENADU - PRATIBHA

EAMCET GRAND TEST (MEDICAL)

The questions of this paper were set by Senior Faculty Members ofSri Chaitanya EAMCET Coaching Centre, Vijayawada.

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1. Study the following lists. ãH÷Ok W=Þ|_# ¬\÷H#° J^μǰ#O KÍÇò=ò

List - I ¬\÷H- I List - II ¬\÷H - IIA) 19th century I) Taxonomy and physiology 19= ×`|íO =s¾H~¡ §¢¬ëO =°iÇò ×s~¡ ^Î~¡à §¢¬ëO

B) 17th century II) Discovery of cell 17= × |íO H Pq+¬ø~¡

C) 18th century III) Role of chromosomes in heredity 18= × |íO P#°=OtH ÇÕ ãHË"³¶*Õ"£° ã Ç

D) 20th century IV) Laws of inheritance 20 = × |íO P#°=OtH ¬¶ã`°

V) Genetic engineering studies *ÿx\÷H± WO[hiOQ· J μǰ<°

The correct match is Wk ¬i³Ø°# *Õ_O¬ô

A B C D A B C D1) IV II I III 2) IV I III V3) I II III IV 4) V II III I

2. Study the following table : – ãH÷Ok ¬\÷H#° KÇ^ΰ=ô=ò

Plant Nature of modified root Host tissue to which it is attached"³òHø ~¡¶O Ç~¡O K³Ok# "Í~¡° ¬Þì=O Wk Ju Í~ò ³òHø D H*ìO`Ë

¬O|O ¥xß U~¡æ~¡KÇ°#°

I) Cuscuta Adventitious Hadrome and Leptome H¬°ø ¼@ J|°Ä~¡¬ô ¬ðã_Ë"£° =°iÇò ÿé"£°

II) Loranthus Normal Hadrome only ç~O ι ª ¥~¡ ¬ðã_Ë"£° =¶ã Ç"Í°

III) Balanophora Adventitious Xylem and phloem ÿ<˦Ú~¡ J|°Ä~¡¬ô ^¥~¡°=ô =°iÇò é+¬H H*ìO

IV) Santalum normal and adventitious Leptome only ªO\ì"£° ª ¥~¡ =°iÇò J|°Ä~¡¬ô ÿé"£° =¶ã Ç"Í°

Select the correct combinations ¬i³Ø°# "Í°×qO¬ô#° Z#°ßHù#°=ò

1) II and III 2) III and IV 3) I and II 4) I and III3. Underground stem with dimorphic roots is kÞ~¡¶¬H "Í×ä Hey# ¶Q®~¡Ä HêO_¡O

1) Rhizome Hù=òà 2) Corm HO ÎO

3) Stem tuber ΰO¬ HêO_ÈO 4) Tuberous stem ^ΰO¬ =O\÷ HêO_ÈO

4. Assertion (A) : In a typical ecosystem insectivorous plants belong to primary producers and second-ary consumersÎ$_È "¼Y¼ (A) : =¶ki P=~¡=¼=¬Ö#O ΰ H©@Hê¬ð~¡ "³òHø° ã Îq°H L Çæuë ¥~¡°° =°iÇò kÞfÇ° qx³¶Q® ¥~¡°°ä½

K³O ΰ#°

Reason (R) : Insectivorous plants are chlorophyllous but feed on insects for nitrogenous sub-stancesHê~¡=ò (R) : H©@Hê¬ð~¡ "³òHø° ¬ìi ÇÇò`° Hêh #ã Ç[x ¬"Í°à×< HˬO H©@Hê#° P¬ð~¡OQê f¬°ä½O\ì~ò

1) Both A and R are true and R explains A A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡

2) Both A and R are true but R doesn't explain A A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡ Hê^ΰ

3) A is true but R is false A ¬ Ǽ=ò, Hêx R J¬ Ǽ=ò

4) A is false but R is true A J¬ Ǽ=ò, Hêx R ¬`Ǽ=ò

BOTANY

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5. Inflorescence of Mangifera differs from that of Oryza in

=¶Ol̦~Õx ¬ô+¬æ q<¼¬O, Xï~á*ìÕx ¬ô+¬æ q<¼¬O #°O_ nxÕ qèk¬°ëOk

I) Mode of arrangement of branches §Y° J=°i=ô#ß q ¥#O

II) Having male and female flowers in addition to bisexual flowers

kÞeOQ® ¬ôæ`Ë @°, ¬ô~¡°+¬ =°iÇò ¢Ôë ¬ôæ#° Hey =ôO_È°@

III) Direction of maturation of flowers ¬ôæ° ¬iu K³O^Í k×

IV) Having pedicellate flowers =$O Ç ¬²ì Ç ¬ôæ#° Hey =ôO_È°@

1) I and III 2) II and IV 3) III and IV 4) I and II

6. Types of placentations found in multicarpellary syncarpous, Multilocular ovaries are

|¬¦¬^Î×Çò`Ç, ¬OÇòHë, |¬aÇò`Ç JO_¨×Ç¶Õ HxæOKÍ JO_È<¼¬ ~¡Hê°

I) Free central ÀÞKó öHOã^Î¬Ö II) Superficial L¬i`Ç

III) Parietal ä½_ȼ IV) Axile JH©ÆÇ°

1) III & IV 2) II & III 3) II & IV 4) I & IV

7. Self pollination does not occur in g\÷Õ P`Çà¬~Q® ¬O¬~¡øO [~¡Q®^ΰ

1) Leafy climber of Fabaceae ¦èÔH÷ K³Ok# ¬ã`Ç #°efQ®`Ë ZQ®ìöH "³òHø

2) Solanaceae member having septifragal capsule and straight embryo

ªÚ<ÍÔH÷ K³Ok, ¬@è^ÎH Q®°oH =°iÇò x~¡ÞãH ²O_ÈO Hey# "³òHø

3) Plants of vernonia, and yucca grown in insect free environment

H©@H ~¡²ì Ç P=~¡Õ Ì~¡°Q®° Ç°#ß "³~ËßxÇ° =°iÇò ÇòHêø "³òHø°

4) A Lilliaceae member bearing unisexual flowers on the clodophylls

eeÍ°ÔH÷ K³Ok, HêÁ_˦² g°^Î UHeOQ®H ¬ôæ° Hey# "³òHø

8. Assertion (A) : Seeds of podostemonaceae and orchidaceae plants are non-endospermic

^Î$_È"¼Y¼ (A) : Ú_ù²"³ò<Ͳ =°iÇò Piø_Ͳ ä½@°O| "³òHø q`Çë<° JOä½~¡KÇó^Î~¡²ì`°.

Reason (R) : The endosperm is completely consumed by the developing embryo during maturity of

seeds of the plants of podostemonaceae and orchidaceae.

Hê~¡=ò (R) : Ú_ù²"³ò<Ͳ =°iÇò Piø_Ͳ "³òHø q`Çë<° ¬HÞ²ÖuH÷ KÍ~¡°#¬C_È° JOä½~¡KÇó^Î=ò#O`ÇÇò

Z^ΰQ®°KÇ°#ß ²O_ÈO ¬ìiOz"ÍÇò#°.

1) Both A and R are true and R explains A

A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡

2) Both A and R are true but R doesn't explain A

A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡ Hê^ΰ

3) A is true but R is false A ¬ Ǽ=ò, Hêx R J¬ Ǽ=ò

4) A is false but R is true A J¬ Ǽ=ò, Hêx R ¬`Ǽ=ò

9. In the Bentham and Hooker's system importance is given to the perianth also in the classification of

ÿO ¥"£° =°iÇò ¬H~ =s¾H~¡ =¼=¬Ö#O ΰ nx =s¾H~¡Õ ¬i¬ã`=oH÷ ä_¨ ã=òY¼ Ç WKó~¡°

I) Monocots UH^Î× c*ì° II) Dicots kÞ^Î× c*ì°

III) Monochlamydae "³¶<ËHêÁq°_Í IV) Gymnospermae q=$ Ç c*ì°

I) I, II, III 2) I and II only I =°iÇò II =¶ã Ç"Í°

3) II and IV only II =°iÇò IV =¶ã Ç"Í° 4) IV alone IV =¶ã Ç"Í°

10. The ratio between the number of microsporophylls in a flower of Tephrosia, megasporophylls in a

flower of withania, microsporophylls in a flower of Yucca and megasporophylls in a flower of

Helianthus is

>ÿã¦é+²Ç¶ ¬ô+¬æO #O^Îe ¬¶HÆ à²^Îc*ì×Ç° ¬ã`°, q^¥xǶ ¬ô+¬æO #O^Îe ¬¶Ö²^Îc*ì×Ç° ¬ã`°, ÇòHêø

¬ô+¬æO #O^Îe ¬¶HÆ à ²^Îc*ì×Ç° ¬ã`° =°iÇò ÔìeǶO^ι ¬ô+¬æO #O^Îe ¬¶Ö ²^Îc*ì×Ç° ¬ã` ¬OY¼

=°^μ x+¬æuë

1) 3 : 5 : 1 : 1 2) 5 : 1 : 3 : 1 3) 5 : 1 : 3 : 4 4) 3 : 1 : 1 : 5

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11. In which group of plants, the number of locules is equal to the number of carpels ?

¦¬^Îì ¬OY¼ä½ aì ¬OY¼ ¬=¶#OQê Hey# "³òHø ¬=ü¬ìO

1) Pisum, Asparagus, Nicotiana and HelianthusÌá¬O, P¬æ~Q®¹, xHË\÷Ƕ# =°iÇò ÔìeǶO^ι

2) Hibiscus, Solanum, Allium and Cucurbita Ìáìa¬ø¹, ªÚì#"£°, PeÇ°O =°iÇò ä½ä½iÄ@

3) Gloriosa, Hibiscus, Crotalaria and solanumQËÁi³¶¬, Ìáìa¬ø¹, ãHË@èiǶ =°iÇò ªÚì#"£°

4) Nicotiana, Lilium, Abutilon and capsicum xHË\÷Ƕ#, eeÇ°O, J|°¼\÷ì<£ =°iÇò Hê²H"£°

12. Study the following table. DãH÷Ok ¬\÷H#° J^μǰ#O KÍÇò=ò.

Cell Organelle Abundantly present in Type of metabolism

HìOQ®=ò JkH ¬OY¼Õ LO_È°#k r=#ãH÷Ç° ~¡HO

I) Mitochondria Meristematic cells Anabolism

"³°Ø\ÕMìOã_Ƕ qì[¼Hì° ¬OõÁ+¬ãH÷Ç°

II) Ribosomes Meristematic cells Anabolism

ï~áÕªé=ò° qì[¼Hì° x~àì`ÇàHãH÷Ç°

III) Peroxysomes Mesophyll cells of C4 plants Amphibolism

Ì~H÷ªé=ò° C4 "³òHø ¬ã`O Ç~¡ Hì° ÎÞO ÎÞr=#ãH÷Ç°

IV) Glyoxysomes Germinating oil storing seeds Catabolism

ïQÁåPH÷ªé=ò° "³òïH Ç°ëKÇ°#ß ãHù=ôÞ#° xÞKÍÇò q Çë<° qKÇó#ßH~¡ãH÷Ç°

The correct combinations are ¬i³Ø°# ä~¡°æ°

1) I Ê II 2) III Ê IV 3) I Ê III 4) II Ê IV

13. The ratio between the number of octomers of histones and that of nucleotides of DNA in a

nucleosome is

XH #¶¼H÷Á³¶ªé=ò#O Îe ²ìªé#° PHË=°~ ¬OY¼ä½ =°iÇò DNA #¶¼H÷Á³¶>ÿØ_£ ¬OY¼ä½ =°^μ x+¬æuë

1) 1 : 20 2) 1 : many (J<ÍHO) 3) 1 : 2 4) 1 : 40

14. If a 10 coiled double helical DNA molecule has 20% adenine in it. then the total number of hydrogen

bonds present in this molecule is

10 "³°eH° Hey# kÞ¬iæ DNA J°=ôÕ J_h<£ 20% L#ß Z_È, P J°=ôÕx "³ò ÇëO L Î[x |O ¥ ¬OY¼

ZO Ç?

1) 100 2) 230 3) 240 4) 260

15. Read the following listsãH÷Ok ¬\÷H#° KÇ^ΰ=ô=ò

List - I ¬\÷H - I List - II ¬\÷H - II

A) Petioles of water lily I) Macrosclereids

h\÷ ("@~) ebÁÕx ¬ã Ç=$O ÇO ¬¶Ö Î$_È Hì°

B) Leaves of Hakea II) Astrosclereids

¬ðH÷ǶÕx ¬ã`° #HÆã`Hê~¡ Î$_È Hì°

C) Testa of Dolichos seed III) Brachysclereids

_¨eH¹ q Çë#OÕx >ÿª ¬¶HÆ à Î$_È Hì°

D) Aerial roots of Monstera IV) Trichosclereids

=¶<£Ì~Õx "ÇòQ® Ç "Í×ä ~Ë=¶Hê~¡ Î$_È Hì°

V) Osteosclereids

J²Ö ^Î$_È Hì°

A B C D A B C D

1) II V I IV 2) V IV II I

3) IV II V III 4) III II IV V

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16. Assertion (A): In Maize stem bundle sheath cells show endodermal nature^Î$_È"¼Y¼ (A): "³òHø*ç#ß HêO_ÈOÕ <oHê¬ôO[ `ù_È°Q®° Hì° JO Ç×ó~¡à Hì ¬Þì=O KǶ¬ô`~ò

Reason (R): Bundle sheath cells of grass leaf possess casparian strips on their radial and transverse wallsHê~¡=ò (R): Q®_Û ¬ã ÇOÕ <oHê¬ôO[ ù_È°Q®° Hì° "¼ª~¡ =°iÇò J_È°Û QË_ÈÌá HêÀæiÇ°<£ =°O^¥#° Hey =ôO\ì~ò

1) Both A and R are true and R explains A A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡

2) Both A and R are true but R doesn't explain A A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡ Hê^ΰ

3) A is true but R is false A ¬ Ǽ=ò, Hêx R J¬`Ǽ=ò 4) A is false but R is true A J¬ Ǽ=ò, Hêx R ¬`Ǽ=ò

17. In a secondary dicot stem the position of recently formed heart wood iskÞfÇ° =$k [iy# kÞ^Î×c[ HêO_ÈOÕ W\©=<Í U~¡æ_# JO`Ç~í~¡°=ô ªÖ#O

1) Just inner to recently formed sap wood W\©= U~¡æ_# ~¡¬^¥~¡°=ôH÷ "³O@<Í Õ¬

2) Just inner to innermost sap wood Õ¬eQê =ô#ß ~¡¬ ¥~¡°=ôH÷ "³O@<Í Õ¬

3) Just inner to vascular cambium <oHê qì[¼ Hì=oH÷ "³O@<Í Õ¬

4) Just inner to early formed heart wood =òO^ΰQê U~¡æ_# JO`Ç~~¡°=ôH÷ "³O@<Í Õ¬

18. Arrange the following in correct ascending order.DãH÷Ok "xx ¬i³Ø°# P~ˬìãH=°OÕ J=°~¡°ó=ò.

I) Number of F2 progeny of dihybrid cross with parental genotypes

[#H [#°¼~¡¶#° Hey# kÞ¬OH~¡¬OH~¡ F2 ¬O`Çu ¬OY¼

II) Number of F2 progeny of dihybrid cross with recombinant phenotypes

¬ô#¬O³¶[H ^Î$×¼~¡¶#° Hey# kÞ¬OH~¡ ¬OH~¡ ¬O`Çu ¬OY¼

III) Number of genotypes of progeny of dihybrid test cross kÞ¬OH~¡ ¬sHÆ ¬OH~¡ ¬O`Çu ³òHø [#°¼ ~¡¶ ¬OY¼

IV) Number of phenotypes of F1 progeny of dihybrid cross

kÞ¬OH~¡ ¬OH~¡ F1 ¬O`Çu ^Î$×¼ ~¡¶ ¬OY¼

1) II, III, I, IV 2) IV, I, III, II 3) IV, III, II, I 4) IV, II, III, I19. Which among the following have same number of chromosomes.

DãH÷Ok "xÕ XöH ãHË"³¶ªé=ò ¬OY¼#° Hey#q.

A) Nullisomic Allium #eÁªéq°H± PeÇ°O

B) Tetrasomic Nicotiana >ÿã\ìªéq°H± xHË\÷Ƕ<

C) Nullisomic pisum #eÁªéq°H± Ìá¬"£°

D) Nullisomic Gossypium #eÁªéq°H± Q겲ǰO

1) B, D 2) A, C 3) A, B 4) C, D20. Read the following lists ãH÷Ok ¬\÷H#° KÇ^ΰ=ô=ò

List - I ¬\÷H - I List - II ¬\÷H - IIA) Wolffia and ceratophyllum I) Rooted plants with floating leaves LeæÇ° =°iÇò Ì~¡\Õ¦²ÁO wßH~¡ K³Ok h\÷Ìá `Íè ¬ã`° Q® "³òHø°

B) Vallisneria II) Free floating plants "e¹<ÍiǶ ÀÞKÇóQê `ÍeǶ_Í "³òHø°

C) Typha and Limnophila III) Submerged and rooted plants >ÿئ =°iÇò e"³¶ß¦² h\÷Õ =òxy =ôO_, wßH~¡ K³Ok# "³òHø°

D) Victoria and Nymphaea IV) Amphibious plants qHËiÇ° =°iÇò xO¦²Ç° LÇ°KÇ~¡ "³òHø°

V) Root less plants "Í~¡°~¡²ì Ç "³òHø°

The correct match is Wk ¬i³Ø°# *Õ_O¬ô

A B C D A B C D1) I III V IV 2) V III I IV3) V III IV I 4) I IV V III

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21. Read the following statements and choose the correct one

ãH÷O^Î Wzó# "¼Y¼#° KÇkq Ç°^¥~¡Ö"³°Ø# "\÷x Z#°ßHù#°=ò

I) Most advanced tracheophytes are non-archegoniates

ìQê ¬iuK³Ok# ã\ìH÷³¶Ì¦á\° PiøQËxǶ#° Hey=ôO_È=ô

II) Most Primitive embryophytes are atracheophytes

ìQê Pk=°"³°Ø# ²O_¨<Íß~¡æiKÍ "³òHø° <oHê H*ìì#° Hey=ôO_È=ô

III) Most advanced archegoniates are primitive Spermatophytes

ìQê ¬iu K³Ok# PiøQËxÍ°\° Pk=°"³°Ø# q Çë# ¥~¡ "³òHø°

IV) Most primitive archegoniates are primitive embryophytes

ìQê Pk=°"³°Ø# PiøQËxǶ<Í~¡æiKÍ "³òHø° Pk=°"³°Ø# ²O_Ë`Çæuë KÍÀ "³òHø°

1) II, III and IV only II, III =°iÇò IV =¶ã Ç"Í° 2) III and IV only III =°iÇò IV =¶ã Ç"Í°

3) II and III only II =°iÇò III =¶ã Ç"Í° 4) I, II, III and IV I, II, III =°iÇò IV

22. In isogamous scalariform conjugation of Spirogyra, the number of empty cells is equal to

Ìåæ~ËïQá~Õx ¬=°¬O³¶Q® *ì`Ç°Õ qãõ÷~¡¶¬H ¬OÇòQ®àO [iöQ@¬C_È° Mìm Hì ¬OY¼ g\÷H÷ ¬=¶#O

1) Number of conjugation tubes ¬OÇòQ®à <ì ¬OY¼

2) Number of zygospores ¬OÇòHë ²^Îc*ì ¬OY¼

3) Total no.of conjugating cells ¬OÇòQ®àOÕ ç¾<Í "³ò ÇëO Hì ¬OY¼

4) Number of cells in one filament XH ÇO Ç°=ôÕx Hì ¬OY¼

23. In Rhizopus stolonifer if the + and – fusing gametangia possess 50 and 70 haploid nuclei respectively

the total number of germspores produced is

ï~á*Õ¬¹ ªÚìx̦~Õ ¬O³¶Q®OÕ ç¾<Í + =°iÇò – ¬O³¶Q® c*ì×Ƕ° =~¡°¬Qê 50 =°iÇò 70 UH²ÖuH

öHOã^ÎHê#° HeyÇò#ß ³°_È, L`Çæuë JÍ°¼ "³ò`ÇëO c[²^Îc*ì ¬OY¼

1) 50 2) 70 3) 120 4) 200

24. In Funaria oblique septa are found in ¦¬ô¼<ÍiÇ¶Õ U@"° ¬\ì#° Hey=ôO_È°#q.

A) Rhizoids =ü ÇO Ç°=ô° B) Trabeculae ã\ìaä½¼è

C) Rhizoidal branches =ü ÇO Ç° §Y°

D) Some changed rhizoidal branches into chloronemal branches

Hùxß ¬ìi Ç ÇO Ç° §Y°Qê =¶i# =ü ÇO Ç° §YÕ

1) All Jxß 2) All Except B B Hêä½O_¨ q°ye#q Jxß

3) All Except C C Hêä½O_¨ q°ye#q Jxß 4) All Except D D Hêä½O_¨ q°ye#q Jxß

25. Following are the structures produced during different stages of development in Pteris

>ÿi¹ Ja=$kÕx qa#ß ^Î×Õ U~¡æ_Í x~àì° ãH÷O^Î W=Þ|_#q

A) Sporophyte ²^Î c[^ÎO B) Sporophyll ²^Îc*ì×Ç° ¬ã`ÇO

C) Sporangium ² Îc*ì×Ç°O D) Embryo ²O_ÈO

E) Gametangium ¬O³¶Q® c*ì×Ç°O F) Gametophyte ¬O³¶Q®c[ ÎO

G) Gametes ¬O³¶Q® c*ì° H) Spore mother cells ² Îc[=¶ Ç$Hì°

Identify the correct sequence of structures produced in between first and last cells of sporophyte

²^Îc[^Î ^Î×Õx "³ò^Î\÷ =°iÇò z=i Hì =°^μ U~¡æ_Í x~àì ¬ï~á# =~¡¬ ãH=¶xß Q®°iëOKÇO_

1) F, E, G 2) D, A, B, C 3) A, B, C, H 4) A, B, C

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26. Select a pair of chracters in which the former is a Pteridophytic character in cycas and latter is an

angiospermic character in Cycas

"³ò^Î\÷ HÆO >ÿi_Ë̦á\H÷ ¬O|OkOz, ï~O_È= HÆO P=$`Çc*ìH÷ ¬O|OkOz, ÌáH¹Õ =ôO_Í [O@ HÆì#°

Q®°iëOKÇO_

1) Presence of independent sporophyte and presence of ovule

¬Þ ÇOã Ç ² Îc[ ÎO =ôO_È°@ =°iÇò JO_ÈO =ôO_È°@

2) Presence of independent sporophyte and presence of haploid endosperm in seeds

¬Þ`ÇOã`Ç ²^Îc[^ÎO =ôO_È°@ =°iÇò q`Çë<Õ UH²ÖuH JOä½~¡KÇó^ÎO =ôO_È°@

3) Presence of ovule and presence of motile Sperms

JO_ÈO =ôO_È°@ =°iÇò KÇ# ¬ô~¡°+¬ c*ì°O_È°@

4) Presence of motile sperms and presence of archegonia inside the ovule

KÇ# ¬ô~¡°+¬ c*ì°O_È°@ =°iÇò JO_ÈOÕ ¢Ôë ¬O³¶Q®c*ì×Ƕ° LO_È°@

27. Assertion (A): In the Bacterial conjugation F– cell becomes F+ cell

^Î$_È "¼Y¼ (A): ìH©iǶ ¬OÇòQ®àOÕ F– HO F+

HOQê ~¡¶ÚO ΰ Ç°Ok

Reason (R): F+ cell receives 'F' plasmid from 'F– cell during conjugation

Hê~¡=ò (R): ¬OÇòQ®àO#O^ΰ F– HO #°O_ 'F' Á²à_£x F+ HO ãQ®²ì¬°ëOk

1) Both A and R are true and R explains A

A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡

2) Both A and R are true but R doesn't explain A

A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡ Hê^ΰ

3) A is true but R is false A ¬ Ǽ=ò, Hêx R J¬ Ǽ=ò

4) A is false but R is true A J¬ Ǽ=ò, Hêx R ¬`Ǽ=ò

28. Choose the correct descending sequence with respect to TMV

TMV H÷ ¬O|OkOz, ¬i³Ø°# J=~ˬìH ãH=¶xß Q®°iëOKÇO_

I) Number of Nuclotides in RNA RNA Õ #¶¼H÷Á³¶>ÿØ_ÈÁ ¬OY¼

II) Number of Strands in RNA RNA Õx éKÇ ¬OY¼

III) Number of Capsomers in capsid Hê²_£Õx Hêéq°Ç°~ ¬OY¼

IV) Number of aminoacids in a capsomer Hêéq°Ç°~Õx J"³°Ø<Ë P=¶Á ¬OY¼

1) III, I, IV, II 2) II, IV, III, I 3) I, III, IV, II 4) II, IV, I, III

29. When cell of 'A' with Q = –0.9 MPa and P = 0.6 MPa and cell 'B' with Q = – 0.8 MPa adn P = 0.3

MPa are placed side by side, the value of : of the cells at equilibrium is

'A' =°iÇò 'B' J<Í ï~O_È° Hì ã¬Høã¬Hø# L<ß~ò. A HOÕ Q = –0.9 MPa =°iÇò P = 0.6 MPa. B

HOÕQ = – 0.8 MPa =°iÇò P = 0.3 MPa ¬=°`²ÖuÕ =ô#߬C_È° P Hì ³òHø : q°=

1) – 0.4 MPa 2) + 0.4 MPa 3) –0.3 MPa 4) –0.1 MPa

30. Identify the common biochemicals involved in the reactions of EMP pathway and PCR cycle

EMP ¬^ÎO =°iÇò PCR =Ç°OÕx KÇ~¡¼ä½ ¬O|OkOz# L=°à_ r=~¡ªÇ°<#° Q®°iëOKÇO_

I) PEP II) Fructose 1, 6 - bis P 㦬HË* 1, 6 - a¹ P

III) DHAP IV) 2-PGA

1) II and III only II =°iÇò III =¶ã Ç"Í° 2) I and III only I =°iÇò III =¶ã Ç"Í°

3) I and IV only I =°iÇò IV =¶ã Ç"Í° 4) II and IV only II =°iÇò IV =¶ã Ç"Í°

(8)

31. Match the following lists with reference to the photolysis of 2H2O

molecules during light reaction of

photosynthesis

H÷~¡[#¼ ¬O³¶Q®ãH÷Ç°Õx HêOu KÇ~¡¼Õ 2 J°=ô H2O HêOu qKÍó Î#O K³O ΰ@ä½ ¬O|OkOz DãH÷Ok ¬\÷H#°

[ Ǭ~¡°KÇ°=ò.

List - I ¬\÷H - I List - II ¬\÷H - II

A) No.of O2 molecules released I) 8

q_È°^ÎÍ°¼ O2 J°=ô ¬OY¼

B) No.of NADPH + H+ produced II) 4

U~¡æ_Í NADPH + H+ ¬OY¼

C) No.of ATP molecules produced III) 12

U~¡æ_Í ATP J°=ô ¬OY¼

D) No.of H+ translocated from stroma to lumen IV) 2

P=iâH #°OKÍ ¶¼"³°<£ÕH÷ ~¡"ì K³O Í H+ ¬OY¼

V) 1

A B C D A B C D

1) II I III IV 2) V IV II I

3) III II I IV 4) V III IV I

32. Number of Oxidations and no.of reduced 'H' acceptors formed respectively in the 3rd step of aerobic

respiration for a glucose molecule is

XH Q®¶ÁHË* J°=ô "Çò§Þ¬ãH÷Ç°Õ ç¾#߬ô_È° =ü_È= Î×Õx PH©H~¡ ¬OY¼ =°iÇò U~¡æ_È° HÆÇ°H~¡

K³Ok# H– ãQ®Ôì`Ç ¬OY¼ =~¡°¬Qê

1) 3 & 6 2) 5 & 10 3) 6 & 12 4) 4 & 8

33. Number of ATP and GTP required for the synthesis of polypeptide chain with 100 aminoacids

100 J"³°Ø<ËP=¶Á`Ë bÌÌå_£ ¬OõÁ+²OKÇ|_È°@ä½ Hê=²# ATP =°iÇò GTP ¬OY¼

1) 100 ATP & 200 GTP 2) 100 ATP & 100 GTP

3) 100 ATP & 199 GTP 4) 99 ATP & 199 GTP

34. Assertion (A) : Plants growing in Zinc deficit soils show normal growth.

Assertion (A) : lOä½ Õ²Oz# <ÍO^ΰ Ì~¡°Q®° "³òHø° ª^¥~¡ suÕ Ì~¡°Q®°#°.

Reason (R) : Zinc is necessary for the synthesis of Tryptophan which is the precursor of IAA.

Reason (R) : IAA ¬î~¡ÞQêq° J~ò# ã\÷é¦<£ ÇǶsH÷ lOä½ J=¬~¡O.

1) Both A and R are true and R explains A

A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡

2) Both A and R are true but R doesn't explain A

A =°iÇò R ¬i³Ø°#q, R J#°#k A ä½ ¬i³Ø°# q=~¡ Hê^ΰ

3) A is true but R is false A ¬ Ǽ=ò, Hêx R J¬ Ǽ=ò

4) A is false but R is true A J¬ Ǽ=ò, Hêx R ¬`Ǽ=ò

35. Methods of crop improvement which do not involve change in the genotype of varieties

~¡HêÕx [#°¼~¡¶¬OÕ =¶~¡°æ [~¡Q®x "³òHø ã¬[## ¬^Î`Ç°°

I) Pureline selection ×à Î=O×ãH=° =~¡O II) Mass selection q§=~¡O

III) Clonal selection HËÁ# =~¡O IV) Introduction ¬ô~¡ªÖ¬#

1) I & IV 2) II & III 3) III & IV 4) I & II

(9)

36. Choose the correct Palindrome sequence from the following.

DãH÷Ok"xÕ Ç°^¥~¡Ö"³°Ø# e<£ã_Ë"£° =~¡°¬ãH=¶xß Q®°iëOKÇO_

1) a a

a a

5 GCC GCC3

3 CGG CGG 5 2) a a

a a

5 ATT ATT3

3 TAA TAA5 3) a a

a a

5 CGG CCG 3

3 GCC GGC5 4) a a

a a

5 GAT TAG3

3 CTA ATC5

37. Identify the correct sequence of events involved in tissue culture experiments

H*ì =~¡Ö#OÕx Î× ³òHø ¬i³Ø°# =~¡°¬ãH=¶xß Q®°iëOKÇO_.

a) Preparation of explant ZH±ÁO\#° ÇǶ~¡° KÍÇò@

b) Incubation for growth Ì~¡°Q®° Î Hù~¡ä½ WOä½¼è+¬<£

c) Preparation of nutrient culture medium é+¬H Ƕ#Hêxß ÇǶ~¡° KÍÇò@

d) Acclimatization of Plantlets ²Á"³òHø#° "`=~¡ìxH÷ J#°ä Ç K³OkOKÇ°@

e) Inoculation of explant ZH±ÁO\#° JO Çiß"Í×#O KÍÇò@

f) Sterilization of nutrient medium =~¡Ö# Ƕ#Hêxß ¬¶HÆ àrq ~¡²ì ÇO KÍÇò@

1) c, f, a, e, b, d 2) c, a, f, e, b, d 3) c, a, e, f, b, d 4) c, d, f, e, b, a

38. Agaricus bisporus is rich in JQêiH¹ ÿáªéæ~¡¹Õ Zä½ø=Qê =ôO_Ík

I) Potassium Ú\ì+²Ç°O II) Phosphorus ¦¬æ~¡¹

III) Iron S~¡<£ IV) Copper Hê¬~ V) Calcium HêeÂÇ°O

1) All except V V Hêä½O_¨ q°ye#=hß 2) All except I I Hêä½O_¨ q°ye#=hß

3) All except III III Hêä½O_¨ q°ye#=hß 4) All except IV IV Hêä½O_¨ q°ye#=hß

39. Pickout the wrong one from following.

DãH÷Ok "xÕ Ç¬CQê L#ß ^¥xx Q®°iëOKÇ°=ò.

1) Cell elongation :- Anti GA property of IAA

H"¼HËKÇ=ò :Ð GA =¼uö~H IAA ^Î~¡àO

2) Induction of seed germination – Anti ABA property of GA

q Çë# JOä½~¡ãÀ~¡ Ð ABA =¼uö~H GA ^Î~¡àO

3) Delay of senescence :- Anti ABA property of cytokinins

"~¡H¼O#° P¬¼O KÍÇò@ Ð ABA =¼uö~H Ìá\ÕïHáx<£ ^Î~¡àO

4) Closure of stomata – Anticytokinin property of ABA

¬ã`Ç~¡Oã^¥° =ü²Hù#°@ Ð Ìá\ÕïHáx<£ =¼uö~H ABA ^Î~¡àO

40. An ideal gene cloning vector must not have

P Î~¡Å=O Ç"³°Ø# [#°¼HËÁxOQ· "³Hê~ qkQê HeyLO_Èä_Èxk

1) Low molecular weight Jæ J°ì~¡O

2) More than one target site for each type of restriction enzyme

ã¬u XH ~¡H"³°Ø# ï~¢²HÆ<£ ZO*ÿá=ò#ä½ XH\÷ HO>è Zä½ø= qt+¬» ¬Öì°

3) 'Ori' site 'Ori' ¬ÖO

4) Drug resistant genes *º+¬^Î x~Ë^ÎH [#°¼=ô°

(10)

ZOOLOGY

41. Find out the codon for aspartate Z¬æ~>è\ä½ ¬OöH`ÇOQê LO_Í Hù_¨<£

1) GAA 2) AAC 3) AAG 4) GAC

42. Study the following ãH÷Ok "xx J^μǰ#O KÍÇò=ò

Bird Distribution Character

¬H÷Æ q¬ë~¡ HÆO

I) Tinamus South Mexico absence of Preen gland

\÷<=°¹ ^ÎH÷Æ "³°H÷HË ãÔ<£ ãQ®Ok èHé=ô@

II) Apteryx New Zealand Absence of aftershaft

ZÌiH± #¶¼lìO_£ J#°²KÇóO Õ²OKÇ°@

III) Rhea South Africa Clavicles are absent

iǶ ^ÎH÷Æ P㦲Hê [ã`Ç°H°O_È=ô

IV) Casuarius New Guinea Absence of keel

Hꬰ"iÇ°¹ #¶¼yxǶ ã^Ë÷ Õ²OKÇ°@

Which of the above are correct ? Ìá "xÕ ¬i³Ø°#q ?

1) Only II and IV II =°iÇò IV =¶ã`Ç"Í° 2) Only II and III II =°iÇò III =¶ã`Ç"Í°

3) Only I, III and IV I, III =°iÇò IV =¶ã`Ç"Í° 4) I, II, III, IV

43. Study the following ãH÷Ok "xx J^μǰ#O KÍÇò=ò

I) Enterokinase acivates chymotrypsinogen ïHá"³¶ã\÷²<Ë[<£#° ZO\÷~ËïHá<Í* L`Íël`ÇO Kͬ°ëOk

II) Leucopenia is caused due to deficiency of folacin ¦éì²<£ Õ¬O =# ¶¼HËÔxǶ H°Q®°`Ç°Ok

III) Muscle weakness is a symptom of hypokalemia HO_È~ |Ôì#`Ç J#°#k ÌáìéHêbq°Ç¶ HÆO

IV) Convulsion is caused due to dificiency of Magnesium "³°wß+²Ç°O Õ¬O =# =ü~¡ó° H°Q®°`~ò

Which of the above are incorrect ? Ìá "xÕ ¬iHêxq ?

1) II, III 2) I, III 3) III, IV 4) I

44. Arrange the following in correct order with respect to the respiratory membrane

§Þ¬`ÇÞKÇOä½ ¬O|OkOz ãH÷Ok "xx ¬i³Ø°# =~¡°¬ãH=°OÕ J=°~¡°ó=ò

a) Capillary basement membrane ~¡HëöH×<oH P^¥~¡`ÇÞKÇO

b) Epithelial basement membrane ¬~¡× ×ø L¬H× P^¥~¡ `ÇÞKÇO

c) Capillary endothelium ~¡HëöH×<oH ZO_ËneÇ°O

d) Alveolar epithelium "ÇòH˧ L¬H×

e) Alveolar air "ÇòHË§Õ Qêe

1) e → b → d → c → a 2) e → b → d → a → c

3) e → d → b → a → c 4) e → d → b → c → a

(11)

45. Match the following ãH÷Ok "xx [`Ǭ~¡KÇ°=ò

A) Bungarus caeruleus 1) Marked with alternate broad, black yellowish rings|OQê~¡¹ Ì~¡¶eÇ°¹ "³_Èÿáæ# #°¬ô, ¬¬°¬ô =Ƕ° UHêO`Ç~¡OQê LO_È°@

B) Ophiophagus hannah 2) Thermoreceptor between nostril and eyeX¦²³¶¦Q®¹ ¬ì<ß <tHê~¡Oã^ÎO =°iÇò H#°ß =°^μ L+¬âãQê¬ìHO LO_È°@

C) Lachesis 3) Arrow mark on headìïH²¹ `ÇÌá "­" Q®°~¡°ë LO_È°@

D) Echis carinatus 4) Hood is with transverse stripesZH÷¹ Hêi<Í@¹ ¬_ÈQ®Ìá J_È°ÛK~¡°O_È°@

5) All subcaudals are in two rowsJxß J^ˬôpóÇ° ¦¬Hê° ï~O_È° =~¡°¬Õ LO_È°@

6) Dorsal surface is bluish or brownish black with narrowwhite cross streaks¬$+¬`ÇO he è^¥ QË^ΰ=° #°¬ô =~¡âOÕ, ³Áx J_È°ÛK~¡`Ë

LO_È°@

A B C D A B C D1) 2 1 3 4 2) 6 4 3 23) 6 4 2 3 4) 1 4 2 3

46. Branchiostoma is characterised by ãìOH÷³¶ªé=¶ HÆO

1) Absence of atrium Uã\÷Ç°O Õ²OKÇ°@

2) Absence of coelom ×s~¡ 佬ì~¡O èHé=ô@

3) Absence of endostyle ZO_ËÌå Õ²OKÇ°@

4) Presence of many pairs of gonads J<ÍH [ Ç c[H˧° LO_È°@

47. Regarding reproductive isolations, structural difference in the genital organs prevent copulation inã¬`Ç°¼`Çæuë q=Hë`Çä½ ¬O|OkOz, ^ÍxÇ°O^ΰ r=ô ã¬`Ç°¼`Çæuë J=Ç°" x~àOÕ =¼`¼ª =#

¬O¬~¡øO x~ËkOKÇ|_È°`Ç°Ok

1) Mechanical isolation 2) Behavioural isolation 3) Temporal isolation 4) Gametic isolation ǶOãuH q=Hë Ç ã¬=~¡ë< q=Hë Ç ¬=°³¶z Ç q=Hë Ç c[Hì q=Hë Ç

48. Sympathetic stimulation ¬¬ð#°¶`Ç ãÀ~¡

1) Constricts coronary arterioles ¬ì$^ÎÇ° ^Î=°xH#° ä½OzO¬ Kͬ°ëOk

2) Decreases blood pressure ~¡HëÔ_È<xß Çy¾OKÇ°#°.

3) Constricts pupil HO\÷¬#° ¬OHËKÇO KÍÇò#°.

4) Mainly constricts blood arterioles ~¡Hë ^Î=°xH#° ã¬^¥#OQê ä½OzO¬ Kͬ°ëOk.

49. Read the following about vaccines and find the correct match.ãH÷Ok "xx Q®=°xOz ¬iQê [`ÇKÍÇ°|_# ^¥xx Q®°iëOKÇO_.

Type of vaccine "H÷<£ ~¡Hê° Example L^¥¬ì

A) Toxoid vaccines \ìHê~ò_£ "H÷#°Á 1) Yellow Fever vaccine ¬KÇó[Þ~¡O "H÷<£

B) Sub – unit vaccines¬Çüx\ "H÷#°Á 2) Diphtheria vaccine _ÔëiǶ "H÷<£

C) Inactivated whole agent vaccines 3) HPV vaccine W<H÷"Í>ÿ_£ ¬ìÙU[O\ "H÷#°Á HPV "H÷<£

D) Attenuated whole agent vaccines 4) Flu vaccine J>ÿ#°¼Í°>ÿ_£ ¬ìÙU[O\ "H÷#°Á ¦¬îÁ[Þ~¡O "H÷<£

5) Type B haemophilous influenzae vaccine >ÿع B ²ì"³¶¦²Á¹ W<£¦¬îÁÇ°O*ì "H÷<£

The correct match is ¬i³Ø°# [`Ç

A B C D A B C D1) 2 3 4 5 2) 2 3 4 13) 2 3 1 4 4) 3 2 4 1

(12)

50. Spool like vertebrae are seen in ¬O>ÿHË =O\÷ Hõ~¡°Hê° Q® rq

1) Diplocaulus 2) Eryops 3) Mud eel 4) Congo eel

_éÁHê¹ Wi³¶¹ =°_£D HêOQË D

51. The maximum number of premolars that are present in marsupials is=¶~¡°²Í° Õ JãQ®KÇ~¡ÞHê Q®i+¬ ¬OY¼

1) 12 2) 4 3) 16 4) 8

52. In rabbit, motor impulses to tongue are carried byä½O^Í°Õ KH ã¬KË^Î<#° <°Hä½ KÍ~¡"ÍÇò <_È°°

1) Hypoglossal nerves l¬ìÞ J Ë<_È°°

2) Mandibular, chorda tympani and Hypoglossal nerves J^ˬì#°, Hê~Û\÷Ox =°iÇò l¬ìÞJ Ë<_È°°

3) Lingual, chorda tympani and Hypoglossal nerves l¬ìÞ§Y, Hê~Û\÷Ox =°iÇò l¬ìÞJ Ë<_È°°

4) Chorda tympani and Hypoglossal nerves Hê~Û\ ÷Ox =°iÇò l¬ìÞJ Ë<_È°°

53. In the embryonic development of rabbit, excretory organs are formed fromä½O^Í° ²O_¨a=$kÕ, q¬~¡ûHê=Ç°"° "Í\÷ #°Oz U~¡æ_È`~ò

1) Scleratome ÔÁø~Ë\ÕO 2) Mesomere g°ªéq°Ç°~

3) Hypomere Ìáìéq°Ç°~ 4) Epimere Z²q°Ç°~

54. In rabbit, the second metatarsal articulates with ä½O^Í°Õ ï~O_È= ã¬^¥²ÖH ^ÍxH÷ ¬OkOKÇ|_È°`Ç°Ok

1) Unciform Jx¦O 2) Mesocuneiform q°ªéä¼x¦O

3) Entocuneiform ZO\Õä¼x¦O 4) Cuboid ä¼ì~ò_£

55. Assertion (A): In Drosophila, 'Y' chromosome has no role in the determination of sex^Î$_È"¼Y¼ (A): ã_˪馲ìÕ eOQ®x~í~¡Õ 'Y' ãHË"³¶*Õ"£°ä½ ZìO\÷ ã`Ç LO_È^ΰ

Reason (R): The cross between a triploid female Drosophila and a normal male results in someaneuploid karyotypesHê~¡O (R): ã`ÇÇ°²ÖuH P_Èã_˪馲ì#° ª^¥~¡ =°Q® ã_˪馲ì`Ë ¬O¬~¡ø ¬~¡KÇ@O =# Hùxß Z#°Á~ò_£

Hêi³¶>ÿع° U~¡æ_È`~ò

1) Both A and R are correct. R is not the correct explanation of A. A =°iÇò R ¬ï~á#q, R J#°#k A ä½ ¬ï~á# q=~¡ Hê ΰ

2) Both A and R are correct. R is the correct explanation of A. A =°iÇò R ¬ï~á#q, R J#°#k A ä½ ¬ï~á# q=~¡

3) A is true, but R is false A X¬C, Hêx R `ǬC

4) A is false and R is true A ¬i³Ø°#k Hê^ΰ, Hêx R ¬i³Ø°#k

56. During formation of urine in rabbit, the secretion of urea into renal fluid occurs inä½O^Í°Õ =üã`ÇO U~¡æ_È° ¬=°Ç°OÕ, =$Hø ã^Î=OÕxH÷ ÇüiǶ ZHø_È ã¬qOKÇ|_È°`Ç°Ok.

1) Descending limb and thin part of ascending limb of loop of Henle

ÌìhÁtH¼¬ô J=~˲ì<oH =°iÇò ¬°KÇx QË_È° Q® P~˲ì<oH ìQ®O

2) Descending limb and thick part of ascending limb of loop of Henle

ÌìhÁtH¼¬ô J=~˲ì<oH =°iÇò =°O Î=°~ò# QË_È° Q® P~˲ì<oH ìQ®O

3) Collecting duct. ¬OãQ®¬ì <×O

4) Collecting duct and DCT ¬OãQ®¬ì <×O =°iÇò ^ζ~¡¬Ö<oH

57. In fishery, drift nets are used in =°`Ǽ ¬iã×=° Ç°O^ΰ, ã_¦¹ =#° ^ÍxÕ "_È`~¡°

1) Inshore fishing W<£é~ ¦²+²OQ· 2) Offshore fishing P¦¹é~ ¦²+²OQ·

3) Inland fishing W<£ìO_£ ¦²+²OQ· 4) Brackish water fishingL¬ôæh\÷ HÇ°¼Õ ¦²+²OQ·

(13)

58. Assertion (A): According to Darwinism, survival is not random^Î$_È"¼Y¼ (A): _¨iÞ<£ ²^¥O`ÇO ã¬Hê~¡O, r=ô =°#°Q®_È J<Ík Ƕ^Î$zóHOQê (W#°ª~¡O) [iöQk Hê^ΰ

Reason (R): The individuals which are deemed fit with their inherited characters leave moreoffspring than the unfit organism in a populationHê~¡O (R): XH [<ìÕ ³¶Q®¼`Ç èx r=ô H<ß, ³¶Q®¼`Ç H r=ô° JkH ¬OY¼Õ ²Á r=ô#° L`Çæuë

KͲ, `Ç^¥Þ~ `Ç=° `Ç~ =°#°Q®_È Hù#ªïöQì KǶªë~ò.

1) Both A and R are correct. R is not the correct explanation of A. A =°iÇò R ¬ï~á#q, R J#°#k A ä½ ¬ï~á# q=~¡ Hê ΰ

2) Both A and R are correct. R is the correct explanation of A. A =°iÇò R ¬ï~á#q, R J#°#k A ä½ ¬ï~á# q=~¡

3) A is true, but R is false A X¬C, Hêx R `ǬC 4) A and R are false A =°iÇò R ǬC

59. Find out the total number of protein coding solitary genes in human genome=¶#=ô_ r<Ë"£°Õ ãé\©<£#° ¬OöHuOKÍ XO@i [#°¼=ô ¬OY¼

1) 3 billion 3 aeÇ°#°Á 2) 30, 000 3) 80, 000 4) 15, 000

60. Method that can be used to accurately measure the density of bone in evaluating osteoporosis is

U ¬^Îíu ^¥Þ~ J²Ö ªOã^Î`Ç#° Hzó`ÇOQê `³°¬°Hù#=KÇ°ó#°

1) ECG 2) CAT Scan CAT ªø<£ 3) X ray X ö~ 4) Polygraphy eãQ®¦²

61. Study the following ãH÷Ok "xx J^μǰ#O KÍÇò=ò

Class qìQ®O Character HÆO Example L^¥¬ì~¡

I) Echinoidea Madriporite is aboral Clypeaster

WH÷<~ò_Ƕ ~¡Oã^Φ¬HO ã¬u=òY`ÇOÕ LO@°Ok ïHáÁ²Ç¶¬~

II) Hirudinea Coelomic cavities in each Pontobdella

segment are separated bymesenteries

²ì~¡°_xǶ ã¬f YO_`ÇOÕx ÔÕq°H± O\ÕÿÛìÁ

佬ì~° POã`dz¶[Hê ^¥Þ~

"Í~¡°KÍÇ°|_ LO\ì~ò

III) Gastropoda Absence of left gill Helix

QêãªéÚ_È Z_È=° "³ò¬æ Õ²OKÇ°@ ÌìeH±

IV) Adenophorea Absence of excretory glands Greeffiella

Z_<˦éiǶ q¬~¡ûH ãQ®O^ΰ° Õ²OKÇ°@ ãw¦²³°ìÁ

Which of the above are incorrect ? Ìá "xÕ ¬iHêxq ?

1) only I and IV =¶ã Ç"Í° 2) only II, III and IV =¶ã Ç"Í°

3) only I, III and IV =¶ã Ç"Í° 4) I, II, III, IV

62. Arrange the following in ascending order based on the number of gills"³ò¬æ ¬OY¼#° |\÷ ãH÷Ok "xx P~ˬì ãH=°OÕ J=°~¡°ó=ò

a) Nautilus <\÷¹ b) Pila Ìáì

c) Chaetoderma H©\Õ_³~à d) Lepidopleurus ÿ²_˦¬îÁ~¡¹

e) Neopilina x³¶²ÿá<

1) b → a → c → e → d 2) c → b → a → e → d3) c → b → a → d → e 4) b → c → a → e → d

(14)

63. Match the following with respect to cockroachçkíOHä½ ¬O|OkOz D ãH÷Ok "xx [`Ǭ~¡°KÇ°=ò

A) Intima with chitin ïHá\÷<£ ¬¡²ì`Ç J=ì²x 1) Stipes Ìá²¹

B) Tympanal organs H~¡âèi JOQê° 2) Mentum "³°O@O

C) Palpifer e榬~ 3) Prementum ãÔ"³°O@O

D)Palpiger eæ[~ 4) Anal cerci ÇòOQê°

E) Subgenual organs J^Ër# JOQê° 5) Tibia \÷aǶ

6) Tarsus \ì~¡¹

7) Tracheoles "Çò<oH°

8) Trachea "Çò<ì°

A B C D E A B C D E1) 7 6 2 1 5 2) 8 4 1 3 53) 7 4 1 3 5 4) 8 4 1 2 5

64. Larva of Antheraea paphia feeds on the leaves ofUOniǶ ¦²Ç¶ ³òHø ì~Þ° "Í\÷ Pä½#° uO\ì~ò

1) Recinus communis ï~²#¹ H=ü¼x¹ 2) Machillus bombycina =¶zÁ¹ ìOa²<

3) Morus alba "³¶~¡¹ PìÄ 4) Terminalia tomentosa >ÿià<eǶ \Õ"³°O\Õª

65. One of the following cells of cockroach secrete the wax layer of the epicuticleçkíOHÕ J^¥¼=ì²xH÷ K³Ok# "³°Ø#¬ôÚ~¡#° ã¬qOKÇ° Hì°

1) Trophocytes ã\Õ¦éÌá@°Á 2) Oenocytes D<ËÌá@°Á

3) Mycetocytes "³°Ø²\ÕÌá@°Á 4) Tormogen cells \ì~Ëà[<£ Hì°

66. In Pheretima, the closed sacs of the coelom that present below the stomach are̦i\÷=¶ Õ r~â×ǶxH÷ L^Î~¡`Çì# LO_Í ÔÕq°H±£ H˧°

1) Ovaries ãÔëc[ H˧° 2) Testis =òø°

3) Prostate glands Ï~¡°+¬ ãQ®O^ΰ° 4) Testes sacs =ò+¬øQË°°

67. During locomotion in Pheretima, contraction of the body is caused due to̦i\÷=¶ Q®=°#O ¬=°Ç°OÕ, ^ͬìO ¬OHËKÇO K³O^ΰ@ä½ Hê~¡O

1) Contration of retractor muscles =òä½o Ç HO_È~ ¬OHËKÇO

2) Contration of longitudinal muscles of the body wall ͬì ä½_ȼO Õx PÇ° Ç HO_È~ ¬OHËKÇO

3) Contration of outer circular muscles of the body wall ͬìä½_ȼOÕx "³°¬e =Ç° HO_È~ ¬OHËKÇO

4) Relaxation of protractor muscles Ja=~¡ëh HO_È~ ¬_ÈeH

68. Match the following ãH÷Ok "xx [`Ǭ~¡KÇ°=ò

A) Epididymis 1) Simple cuboidal epitheliumZ²__³áq°¹ ¬~¡× ¦¬°<Hê~¡ L¬H×

B) Urethra 2) Stratified squamous non keratinised epitheliumã¬ÀHO ¬ëi`Ç ×ø ïH~¡\÷<£ ~¡²ì`Ç L¬H×

C) Conjunctiva of eye 3) Pseudostratified non ciliated columnar epitheliumHO\÷ HO*ÿOïHå" q°^¥¼¬ëi`Ç ÜáeHê~¡²ì`Ç ¬ëOìHê~¡ L¬H×

D) Germinal epithelium 4) Simple columnar ciliated epithelium[## L¬H× ÜáeHꬲì`Ç ¬~¡× ¬ëOìHê~¡ L¬H×

E) Vagina ³¶x 5) Transitional epithelium ¬i=~¡ë# L¬H×

6) Stratified columnar epithelium ¬ëi`Ç ¬ëOìHê~¡ L¬H×

7) Stratified cuboidal epithelium ¬ëi`Ç ¦¬°<Hê~¡ L¬H×

A B C D E A B C D E1) 6 3 4 1 7 2) 3 5 7 1 23) 6 3 4 1 2 4) 3 5 6 1 2

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69. Periosteum is made of ¬~¡¼²ÖH ^Íx`Ë xià`Ç=°=ô`Ç°Ok

1) Reticular connective tissue *ìH ¬O³¶[H H*ìO

2) Regular dense fibrous connective tissue ªOãnÇ° ãH=°Çò Ç ¬O³¶[H H*ìO

3) Elastic dense fibrous connective tissue ²ÖuªÖ¬H ¬O³¶[H H*ìO

4) Irregular dense fibrous connective tissue ªOãnÇ° ãH=°~¡²ì`Ç ¬O³¶[H H*ìO

70. Assertion (A) : Genetic recombination occurs in cytogamy in Paramoecium caudatum

~¡g°²Ç°O Hê_Í@"£°Õ [iöQ Ìá\ÕQ®q° Ç°O^ΰ [#°¼¬ô# ¬O³¶[#O [~¡°Q®°`Ç°Ok

Reason (R) : Fusion of pronuclei of two individuals occurs in only a single individual duringcytogamyÌá\ÕQ®q°Õ, ï~O_È° r=ôä½ K³Ok# ãöHøOã ÎHê H~òH XöH XH rqÕ =¶ã Ç"Í° [~¡°Q®° Ç°Ok.

1) Both A and R are correct. R is not the correct explanation of A. A =°iÇò R ¬ï~á#q, R J#°#k A ä½ ¬ï~á# q=~¡ Hê ΰ

2) Both A and R are correct. R is the correct explanation of A. A =°iÇò R ¬ï~á#q, R J#°#k A ä½ ¬ï~á# q=~¡

3) A is true, but R is false A X¬C, Hêx R `ǬC 4) A and R are false A =°iÇò R ǬC

71. In the life cycle of Taenia solium, the bladder gets disintegrated in\©xǶ ªéeÇ°"£° rq`Ç KÇãHOÕ, uuëx HËÕæ=ô@ ZHø_È [~¡°Q®°`Ç°Ok

1) Duodenum of pig ¬Ok POã Ç=üO 2) Intestine of man =¶#=ôx ÀQ®°

3) Voluntary muscles of pig ¬Ok xÇ°OãuO Ç HO_È~° 4) Stomach of man =¶#=ôx r~â×Ç°O

72. An annelid with parapodia is characterised by~¡ÅÞ ^¥° Q® J<³e_£ #O^ΰ D ãH÷Ok HÆìxß Q®°iëOKÇ=KÇ°ó#°

1) Botryoidal tissue Õã\÷Ƕ~ò_È H*ì=ò 2) Direct development ã¬`ǼHÆ ²O_¨a=$k

3) Absence of septa qì[Hê° Õ²OKÇ°@ 4) Absence of clitellum ïHåÁ>ÿÁ"£° Õ²OKÇ°@

73. In Pheretima, the blood vessels that supply bood to micronephridia and the vessel that collectsblood from it are

̦i\÷=¶ #O^ΰ ¬¶HÆ à=$Hêøä½ ~¡Hêëxß ¬~¡¦¬~ KÍÇò =°iÇò "\÷ #°O_ ~¡Hêëxß Qùx é=ô ~¡Hë <ì°

1) Ventrotegumentary vessels and lateral oesophageal vessels

`ÇÞKË^Î~¡ ~¡Hë<ì° =°iÇò ~¡ÅÞ P¬ð~¡"²ìHê ~¡Hë<ì°

2) Dorsal blood vessel, Subneural vessel ¬$+¬ ~¡Hë <×=ò, J^Ë <_ ~¡Hë<×=ò

3) Lateraloesophageal vessel,Ventral blood vessel ~ÅÞ¬ð~¡"²ìH ~¡Hë<×O, L^Î~¡ ~¡Hë <×=ò

4) Dorsal blood vessel, Supraoesophageal blood vessel ¬$+¬ ~¡Hë<×=ò, J ¥¼¬ð~¡ "²ìHê~¡Hë <×=ò

74. Match the following and choose the correct answer

D ãH÷Ok "xx [`Ǭ~¡z ¬i³Ø°# ¬=¶^¥#=ò H#°Qù#°=ò

A) Benthos ÿO^¥¹ I) Gerris *ÿãi¹

B) Neckton <³Hê<£ II) Trionyx ã@³¶xH±

C) Epineuston Z²#¶¼ª<£ III) Astacus Z¬H¹

D) Periphyton Ìi̦á\ì<£ IV) Fresh water snails =°Ozh\÷ #`Çë°

1) A IV, B II, C I, D III 2) A III, B II, C I, D IV

3) A II, B IV, C III, D I 4) A III, B II, C IV, D I

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75. Non chordate with closed type of circulation that show mosaic embryos are characterised by"³ò*ì~òH± ²O_¨#° ã¬^ÎiÅOKÍ ¬O=$`Ç ~¡Hë㬬~¡ Q® Hêö~Û>è`Ç~¡ [O`Ç°=ô HÆO

1) Presence of pseudocoelom q°^¥¼×s~¡ 佬ì~¡O LO_È°@

2) Spiral and determinate type of cleavage ¬iæ =°iÇò x~i Ç q Î×<°

3) Presence of enterocoelom POã`Ç佬ì~¡O LO_È°@

4) Presence of mesodermal skeleton =° μ¬ë ÞKÇ J²Ö¬O[~¡O LO_È°@

76. During life cycle of Plasmodium vivax, the formation of gametes takes place inÁªéà_Ç°"£° "³á"H± rq`Ç KÇãHOÕ, ¬O³¶Q® c*ì° ZHø_È U~¡æ~¡`~ò.

1) Plasma of man =¶#=ôx Áªà 2) Liver of man =¶#=ôx HêèÇ°O

3) Spleen and bone marrow ÔÁ¬ìO =°iÇò Z=òH =°[û 4) Crop of mosquito ^Ë=° J<ß×Ç°O

77. Arrange the following in ascending order based on their number in Indiaì~¡`Ç^Í×OÕx "\÷ ¬OY¼#° |\÷ ãH÷Ok "xx P~ˬì ãH=°OÕ J=°~¡°ó=ò

a) Biosphere reserves r=QË× i[~¡°Þ° b) Ecoregions ºQËoH =°O_Èì°

c) Hot spots ¬ð\ªæ\° d) Sanctuaries ¬O~¡HÆìǶ°

e) National parks *ìfÇ° ~¡°ø°

1) a → b → c → d → e 2) c → a → e → b → d 3) c → a → b → e → d 4) c → b → a → e → d

78. Assertion (A) : In camel, cooling the body is very efficient compared to the amount of water lostthrough sweatingXO>ÿÕ Í¬ìO KÇÁ|_È@O J<Ík K³=°@ ¥Þ~ HËÕæ~ò# h\÷ "³ò`ÇëO`Ë éeó`Í ¬=°~¡í=O`ÇOQê

[~¡°Q®°`Ç°Ok.

Reason (R) : In camel evaporation of sweat takes place at the surface of its coat not at skin levelXO>ÿÕ, K³=°@ Lxß L¬i ÇO Ìá PqiQê =¶~¡° Ç°Ok, Hêx KÇ~¡à ªÖ~ò = Îí Hê ΰ

1) Both A and R are correct. R is not the correct explanation of A. A =°iÇò R ¬ï~á#q, R J#°#k A ä½ ¬ï~á# q=~¡ Hê ΰ

2) A is true, but R is false A X¬C, Hêx R `ǬC 3) A is false, but R is true A ǬC,Hêx R X¬C

4) Both A and R are correct. R is the correct explanation of A. A =°iÇò R ¬ï~á#q, R J#°#k A ä½ ¬ï~á# q=~¡

79. Choose the correct statements with respect to Vorticella=iÌìÁä½ ¬O|OkOz ¬i³Ø°# "¼Y¼#° Q®°iëOKÇ°=ò

I. It undergoes binary fissions during favourable conditions J#°ä ¬i²Ö`Ç°Õ Wk kÞ^¥qzó`Ç°ë° K³O^ΰ`Ç°Ok

II. Microconjugant neither feeds nor encysts¬¶HÆ à¬OÇòQ®àHO P¬ð~¡ÀH~¡ QêqOKÇ_ÈO Qêx HËj=#O K³O^Î_ÈO Qêh [~¡¬è^ΰ

III. Attachment of the two conjugants takes place on their aboral surfacesï~O_È° ¬OÇòQ®àHê H~òH "\÷ ã¬u=òY `ÇìÕ [~¡°Q®°`Ç°Ok

IV. The cytoplasm of conjugants does not divide in prezygotic nuclear divisions¬OÇòHëöHOã^ÎH ¬î~¡Þ q[#ÕÁ Hã^Î=¼O q[# [~¡Q®^ΰ

1) Only III & IV =°ã`Ç"Í° 2) Only II, III, IV =¶ã`Ç"Í° 3) Only II & IV=¶ã`Ç"Í° 4) I, II, III, IV

80. Kingdom that includes multicellular saprobic heterotrophs was created by|¬H ¬îuHê¬ð~¡ ¬~¡é+¬Hê° Q® ''~[¼OÑÑ#° ã¬ukOz#"~¡°

1) Whittaker q>èH~ 2) Copeland H˹ìO_£

3) Haeckel ÌìïH 4) Linnaeus e<ÍßÇ°¹

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PHYSICS

81. List I (¬\÷H) -I List -II (¬\÷H) -II

a) Second ÌHO_£ e) C-12 Hê~¡Ä<£ Ð 12

b) Mole "³¶ f) platinum- irridium Á\÷#"£° Ð Wi_Ç°O

c) Metre g°@~¡° g) Cs-133 ԲǰO Ð 133

d) Kilogram H÷ÕãQê=ò h) Kr-86 ãH÷<£ Ð 86

1) a-e, b-g, c-h, d-f 2) a-f, b-h, c-g, d-e 3) a-e, b-f, c-h, d-g 4) a-g, b-e, c-h, d-f.

82. A stone is projected with a velocity 110 2 ms at an angle of 45º to the horizontal ground. Theaverage velocity of stone during its time of flight is

XH ~~òx H÷Æu[ ¬=¶O Ç~¡ Ç=ò`Ë 450

HË=ò KÍÇòKǶ 110 2 ms "ÍQ®=ò`Ë ã¬H÷Ƭë=ò KͲ<~¡°. J~ò#

¬ìÇ°# Hê=ò #O ΰ P ~~ò ³òHø ¬Q®@° "ÍQ®=ò

1) 110 ms 2) 110 5 ms 3) 15 5 ms 4) 120 ms

83. A hunter fires a gun from a boat. The mass of the bullet is 35g. The velocity of the bullet is 320 m/sand the gun is fired at 600 to horizontal. If mass of the hunter and boat is 70 kg. The recoil velocityof boat is XH "Í@Qê_È° ¬_È= g° Î #°OKùx

XH Ç°H©x H÷Æu[ ¬=¶O Ç~¡=ò`Ë 600

HË=òÕ LOz 35gm ã Î=¼~t Q® |°ÿÁ\#° 320 m/s "ÍQ®=ò`Ë Àeó<~¡°.

"Í@Qê_È° =°iÇò ¬_È= ³òHø ¬"Í°à×< ã^Î=¼~t 70 kg J~ò# ¬_È= ³òHø 㬠¼=~¡ë# "ÍQ®=ò

1)10 m/s 2) 12 m/s 3) 0.08 m/s 4) 0.04 m/s84. A boy of mass 60kg standing on a platform of mass 40kg placed over a smooth horizontal surface.

He throws, a stone of mass 1kg with velocity u = 10 m/s at angle of 450 w.r.t the ground. The distancebetween the boy and stone on the horizontal surface when the stone lands on the ground is(g = 10 m/s2)XH #°#°Ìá# H÷Æu[ ¬=¶O`Ç~¡ `Ç=òÌá# 40kg ã Î=¼~t Q® XH k=°à L#ßk. ^¥xÌá# 60kg ã Î=¼~t Q® XH

²Á"_È° L<ß_È°. J`Ç_È° 1kg ã Î=¼~t Q® XH ~~òx u = 10 m/s "ÍQ®=ò`Ë H÷Æu[ ¬=¶O Ç~¡=ò`Ë 450

HË=ò`Ë

ã¬H÷Ƭë=ò KͲ<_È°. P ~~ò <Í g°^Î ¬_È°¬=°Ç¶xH÷ ²Á"_H÷ ~~òH÷ =°^μ L#ß ^ζ~¡=ò (g = 10 m/s2)1)10 m 2)0.1m 3)10.1m 4)9.9m

85. A vertical spring with force constant k is fixed on a table. A ball of mass m at a height h above thefree upper end of the spring falls vertically on the spring so that the spring is compressed by adistance d. The net work done in the process isK |²Ö~OH=ò Q® XH ã²æOQ®°x x@x°=ôQê XH |Áä½ ayOz<~¡°. P ã²æOQ· ³òHø ï~O_È= z=~¡ #°O_ h Z Ç°ë

#°O_ m ã^Î=¼~t Q® XH |Oux ã²æOQ· g°^Îä½ x@x°=ôQê *ì~¡q_È°KÇ°@ =# P ã²æOQ· d ^ζ~¡O ¬OÔ_È#O

K³Ok#k. J~ò# D "³ò ÇëO ã¬ãH÷Ç°Õ [iy# ¦¬e Ǭx

1) 21

mg(h d) kd2

2) 21

mg(h d) kd2

3) 21

mg(h d) kd2

4) 21

mg(h d) kd2

86. A small spherical ball strikes a frictionless horizontal plane with a velocity 'V' making an angle 'R 'to the normal at the surface. If the coefficient of restitution is 'e', the particle will again strike the

surface after timeXH z#ß QËìHê~¡ |Ou 'V' J<Í "ÍQ®=ò`Ë Ç=ò ³òHø O|=ò`Ë 'R ' J<Í HË=ò KÍÇòKǶ #°#°Ìá# H÷Æu[

¬=¶O Ç~¡ Ç=ò`Ë Ja¦¬¶ Ç=ò [i²#k. 㬠Ǽ=ªÖ# Q®°H=ò 'e', J~ò# Z_È P |Ou =°~¡ì Ç=ò`Ë

Ja¦¬¶`Ç=ò KÍÇò@ä ¬@°Hê=ò

1) 2 sinV

g

R2)

2 coseV

g

R3)

2 sineV

g

R4)

2 cosV

g

R

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87. A block of mass 10kg pushed by a force F on a horizontal rough plane moves with an aceelerationof 5ms–2. When force is doubled, its acceleration becomes 18ms–2. Find the coefficient of friction

between the block and rough horizontal plane. (g = 10ms–2). 10kg ã^Î=¼~t Q® XH

k=°à#° F J<Í |=ò`Ë H÷Æu[ ¬=¶O Ç~¡ Q®~¡°ä½ Ç=òÌá <³\÷#¬ôæ_È° Jk ÚOk# ÇÞ~¡=ò 5ms–2. |=ò#° ï~\÷O¬ô

KͲ#Ç°_È Jk ÚOk# ÇÞ~¡=ò 18ms–2. J~ò# ¦¬°~¡Â Q®°H=ò (g = 10ms–2).1) 0.6 2) 0.8 3) 0.4 4) 0.2

88. A body of mass 5 kg rests on a rough horizontal surface of coefficient of friction 0.2. The body ispulled through a distance of 10 m by a horizontal force of 25 N. Find the kinetic energy acquiredby the body. ( )-2g = 10 ms

5 kg ã Î=¼~t Q® XH =¬°ë=ô 0.2 ¦¬°~¡Â Q®°H=ò Q® XH Q®~¡°ä½ H÷Æu[ ¬=¶O Ç~¡ Ç=ò g° Î q~=°²ÖuÕ L#ßk.

P =¬°ë=ôx 25 N H÷Æu[ ¬=¶O Ç~¡ |=ò`Ë 10m ζ~¡=òìy# Ç°_È P =¬°ë=ô ÚOk# Q®u[ ×H÷ë ( )-2g = 10 ms

1) 50 J 2) 100 J 3) 150 J 4) 200 J

89. A duster weighs 0.5N. It is pressed against a vertical board with a horizontal force of 11N. If theco-efficient of friction is 0.5 the minimum force that must be applied on the duster parallel to theboard to move it upwards is

0.5N ì~¡=ò Q® _Ȭ~#° XH #ÁÕ~¡°Û Ìá 11N H÷Æu[ ¬=¶O Ç~¡ |=ò`Ë Jkq° ¬@°ä½<ß~¡°. N « 0.5 J~ò#

P _Ȭ~#° Õ~¡°ÛH÷ ¬=¶O`Ç~¡=òQê ÌáH÷ H^ΰ°ó@ä½ J^Í k×Õ J=¬~¡=°Q®° |=ò

1) 0.4 N 2) 0.7 N 3) 6 N 4) 7 N

90. Statement A: When a force is applied at centre of mass of a body it may undergo translatorymotionStatement B: At the centre of mass there is always some mass particleStatement A: XH =¬°ë=ô ³òHø ã Î=¼~t öHOã Î=òÌá ì¬ì¼|=ò J#°=iëO¬KͲ#¬ô_È° Jk ªÖ<O Ç~¡ KÇ#=ò#°

ÚO ΰ#°.

Statement B: ã Î=¼~t öHOã Î=ò = Îí ZÁ¬C_È° HùO Ç ã Î=¼~t =ôO_È=ÿ#°.

1) A is true B is false A ¬i³Ø°#k B ¬iHêxk 2) A is false B is true A ¬iHêxk B ¬i³Ø°#k

3) Both A and B are true A =°iÇò B ° ¬i³Ø°#q

4) Both A and B are false A =°iÇò B ° ¬iHêxq

91. A body of radius 'R' and mass 'm' is rolling horizontally without slipping with speed 'V'. It then rolls

up a hill to a maximum height 23v

h4g

= . The body might be a

'm' ã Î=¼~t, 'R' "¼ª~¡Ö=ò Q® XH =¬°ë=ô 'V' "ÍQ®=ò`Ë H÷Æu[ ¬=¶O Ç~¡ ÇOÌá *ì~¡ä½O_¨ ù~¡°ÁKÇ°#ßk. P =¬°ë=ô

^ù~¡°Á`Ƕ XH "° `ÇO ÌáH÷ Zä½ø`Ç°#߬ôæ_È° Jk KÍi# Q®i+¬ Z`Ç°ë

23vh

4g= . P =¬°ë=ô

1) Solid sphere 2) Hollow sphere 3) Disc 4) Ring

¦¬°#QË×=ò Õ° QË×=ò =$`ëHê~¡ aפ LOQ®~¡=ò

92. Moment of inertia of a uniform circular disc about a diametric axis AB that passes through itscentre is I. CD is a line in the plane of disc that passes through the centre of the disc and makesangle 300 with AB. The moment of inetia of the disc about the axis CD is

UHsu =$`ëHê~¡ aפ "¼¬O Q®°O_¨ é Ç°#ß AB JHÆO ¬~¡OQê [_È ÇÞ ãì=°H=ò I. CD J<Í ö~M =$`ëHê~¡ aפ

³òHø Ç=òÕ LO_ P aפ ³òHø öHOã Î=ò #°O_ é=ôKÇ°#ßk. XH "Í× CD =°iÇò AB ä½ =° μ HËO 300

J~ò Í CD J<Í ö~Y Q®°O_¨ é Ç°#ß JHÆ=ò ¬~¡OQê =$`ëHê~¡ aפ ³òHø [_È ÇÞ ãì=°H=ò

1) 2

I2) I

2

3 3) 4

I34) I

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93. The escape velocity of a body on the earth's surface is eV . A body is thrown with a speed e3V . Assum-

ing that the sun and planets do not influence the motion of the body, its speed at infinity would be

¶q° L¬i Ç=ò g° Î XH =¬°ë=ôä½ ¬ìÇ°# "ÍQ®=ò eV . W¬ôæ_È° XH =¬°ë=ô#° e3V "ÍQ®=ò`Ë ã¬H÷Ƭë=ò KÍÀë, J#O Ç

ζ~¡=ò#O ΰ =¬°ë=ô "ÍQ®=ò (¬¶~¡°¼_È°, q°ye# ãQ®¬ð° =¬°ë=ô Q®=°#=ò#° ã¬ìq Ç=ò KÍÇ°=ô J#°Hù#°=ò)

1) zero ¬°#ß 2) eV 3) e2V 4) e2 2V

94. The potential energy of a particle executing SHM in its mean position is 15J. The average kinetic

energy of the particle during one oscillation is 5J. The total energy of the particle is

¬.¬ì.KÇ Õ L#ß H=ò ^¥x ³òHø =¶ μq°HªÖ#=òÕ L#߬ôæ_È° ^¥x ²Öu[ ×H÷ë 15J. XH ¬îië HO¬#=ò

[iy#¬ôæ_È° P H=ò ³òHø ¬Q®@° Q®u[×H÷ë 5J. J~ò Í P H=ò ³òHø "³ò Çë=ò ×H÷ë

1) 10J 2) 25J 3) 15J 4) 5J

95. A body of mass 10kg is attached to a wire 0.3m long. Its breaking stress is 4.8x107 N/m2 . The

area of cross section of the wire is 10-6m2 . The maximum angular velocity with which it can be

rotated in a horizontal circle without breaking is

0.3 m Ú_È=ô Q® fQ®ä½ 10kg ã^Î=¼~t Q® =¬°ë=ô#° ã"Íì_Èn§~¡°. ¥x qKÍó^Î# ã¬u|O 4.8x107 N/m2. fQ®

=°^μKÍó^Î "³á§¼O 10-6m2 J~ò# Jk ³Q®ä½O_¨ ZO Ç Q®i+¬ HË©Ç° "ÍQ®=ò`Ë H÷Æu[ ¬=¶O Ç~¡ =$ ÇëOÕ u~¡Q®Q® ΰ

1) 2rad/s 2) 4rad/s 3) 6rad/s 4) 8rad/s

96. The excess pressure inside a soap bubble is x times the excess pressure inside another soap bubble.

the volume of the first bubble is 0.125 times the volume of the second bubble. Then the value of

x is XH ¬|°Ä |°_ÈQ® Õ¬ JkH Ô_È#=ò ï~O_È= ¬|°Ä |°_ÈQ® Õ¬ JkHÔ_È#=òä½ x ï~@°Á

L#ßk. "³ò Î\÷ ¬|°Ä |°_ÈQ® ³òHø ¦¬°#¬i=¶=ò ï~O_È= |°_ÈQ® ¦¬°#¬i=¶=ò H<ß 0.125 ï~@°Á J~ò# x q°=

1) 1 2) 2 3) 4 4) 8

97. Due to ejection of water (density ρ ) with a velocity v from a nozzle of area of crosssection A, the

spring of spring constant K is compressed. In the equilibrium condition, the compression of the

spring is

ρ J<Í ªOã Î Ç, v J<Í "ÍQ®=ò`Ë A J<Í =°^μKÍó^Î "³á§¼=ò Q® XH ~¡Oã^Î=ò #°O_ h~¡° |Ç°@ä½ ã¬=²ìOKÇ°@

=# K J<Í | ²Ö~OH=ò Q® ã²æOQ· ¬OHËKÇ=ò K³Ok`Í ¬=°` ²Öu Ç°O^ΰ P ã²æOQ·Õ HeöQ ¬OHËKÇ=ò

1) 2

2

v A

K

ρ2)

2v A

K

ρ

3) ρν2AK 4) zero

98. One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm areintroduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of

the gas in atm is

1 b@~¡° =°iÇò 1 atm Ô_È#=ò = Îí L#ß PH÷[<£ #° =°iÇò 2 b@~¡°Á =°iÇò 0.5 atm Ô_È#=ò = Îí L#ß

<³áã\Õ[<£#° He² 1 b@~¡° ã`ÇÕ#H÷ ã¬"Í×Ì\÷<~¡°. LéâãQ®`ÇÕ =¶~¡°æè#¬ôæ_È° P q°ã×=° "Çò=ô ³òHø ¦¬e`Ç

Ô_È#=ò q°=

1) 1.5 2) 2.5 3) 2 4) 4

(20)

99. When 100 J of heat is applied to a thermodynamic system containing mono atomic gas then the

percentage of energy converted into external work done is 53

γ =

UH ¬~¡=¶°H "Çò=ô L#ß XH L+¬âQ®uH =¼=¬Öä½ 100 J ×H÷ë ¬~¡¦¬~ KͲ# Ç°_È ×H÷ëÕ ZO Ç § Ç=ò ¬xKÍÇò@Õ

L¬³¶Q®¬_È°`Ç°Ok.

53

γ = 1) 30 % 2) 40 % 3) 60 % 4) 80 %

100. During an experiment, an ideal gas is found to obey an additional law VP2 = constant, the gas

being initially at temperature T and volume V. When it expands to volume 2V, the temperature

becomes.XH 㬳¶Q®OÕ P^Î~¡Å"Çò=ô VP2 = ²Ö~¡~t J#° J Î#¬ô xÇ°=¶xß \÷OKÇ°#°. ^¥x `ùe LéâãQ® Ç T =°iÇò

¦¬°#¬iì=°=ò V. Jk 2V, ¦¬°#¬iìxH÷ "¼HËzOz#¬ôæ_È° ^¥x LéâãQ®`Ç

1) 2 T 2) 2T 3) 2

T4)

2

T

101. Six identical conducting rods are joined as shown in figure. Points A and D are maintained at tem-peratures 2000C and 200C respectively. The temperature of junction B will be.

P~¡° ¬~¡Þ¬=¶#"³°Ø# "¬ìH H_Û#° ¬@OÕ KǶ²# q Î=òQê He²<~¡°. A =°iÇò B aO ΰ=ô = Îí L#ß LéâãQ® Ç°

=~¡°¬Qê 2000C =°iÇ° 200C J~ò Í B ¬Ok =^Îí L#ß LéâãQ®`Ç

A B C D0200 C 020 C

1) 1200 C 2) 1000 C 3) 1400 C 4) 800 C

102. An auditorium has volume 105 m3 and surface area of absorption 2 x 104 m2. Its average abosorptioncoefficient is 0.2. The reverberation time of the auditorium (in seconds)is105 m3

¦¬°#¬i=¶O, 2 x 104 m2 "³á§¼=ò =°iÇò 0.2 ¬Q®@° Õ+¬ Q®°H=ò Q® ¬ð° ³òHø ã¬u<^Î Hê=ò

1) 6.5 2) 5.5 3) 4.25 4) 3.25

103. A man is standing on the platform and one train is approaching and another train is going away withspeed of 4 m/s, frequency of sound produced by train is 240 Hz. What will be the number of beatsheard by him per second (Take velocity of sound = 320 m/s)

XH =¼H÷ë ¬~¡OQê ï~O_È° ï~á×¤Õ XH\÷ J Çxx ¬g°²OKÇ°KÇ°#ßk. ï~O_ù=k ζ~¡OQê é=ôKÇ°#ßk. XHùøHø ï~á° ³òHø

"ÍQ®O 4 m/s, ã¬uï~á° ³òHø ql Ï#:¬ô#¼O 240 Hz. J~ò# P =¼H÷ë XH ÌH#°Õ q<Í q¬æO^Î< ¬OY¼ (^ÎÞx

"ÍQ®O = 320 m/s)1) 12 2) zero 3) 6 4) 3

104. For an equilateral prism, it is observed that when a ray strikes grazingly at one face it emergesgrazingly at the other. Its refractive index will be

XH ¬=°ì¬ ãu°[ ¬@HO ³òHø XH =ãH©=# `ÇO#° ¬°$t¬¶ë ¬`Ç#=ò° =°iÇò ï~O_È= `Çìxß ¬°$t¬¶ë

|²ì~¡¾ ÇO J~ò Í P ¬@H=ò ³òHø =ãH©=# Q®°H=ò q°=

1) 3 2) 2

33) 2 4) Data not sufficient

(21)

105. A crown glass prism of refracting angle 8o is combined with a flint glass prism to obtain deviationwithout dispersion. If the refractive indices for red and violet rays for crown glass are 1.514 and1.524 and for the flint glass are 1.645 and 1.665 respectively, Then the net deviation is

80 z#ß ¬@H HËO Q® ãHÒ<£ QêA¬@H=ò#ä½ qKÇ#O èä½O_¨ qöHƬO U~¡æ_Í@@°ÁQê ²ÁO\ QêA ¬@H=ò#°

He²<~¡°. ãHÒ<£QêA ¬@H=ò ³òHø =ãH©=# Q®°H=ò° T^¥~¡OQ®°ä½ 1.524 =°iÇò Z~¡°¬ô ~¡OQ®°ä½1.514

=°iÇò ²ÁO\ QêA ¬@H=ò ³òHø =ãH©=# Q®°H=ò° =~¡°¬Qê 1.665 =°iÇò 1.645 J~ò Í ¦¬e Ç qöHƬO

1) 1.530 2) 2.530 3) 4.530 4) 3.530

106. An air bubble is trapped inside a glass cube of edge 30 cm. Looking through the face ABEH, thebubble appears to be at normal distance 12 cm from this face and when seen from the oppositeface CDGF, it appears to be at normal distance 8 cm from CDGF. Then the refractive index ofglass 30 cm °[=ò Q® XH QêA ¦¬°#=òÕ XH Qêe |°_ÈQ® L#ßk. ABEH, =òY=ò #°O_

KǶ²#¬ôæ_È° P |°_ÈQ® P =òY=òä½ 12 cm O| ζ~¡=òÕ L#ß@°Á =°iÇò CDGF =òY=ò "³á¬ô #°O_ KǶ²#¬ôæ_È°

P |°_ÈQ® CDGF =òY=ò#ä½ 8 cm O| ζ~¡=òÕ L#ß@°Á Hx²¬°ëOk. J~ò# P QêA ³òHø =ãH©=# Q®°H=ò

1) 2

A

B C

D

E F

GH

2) 1.5

3) 2.5

4) 1.75

107. A long bar magnet arranged vertically at '10cm' on the eastern arm of D.M.M. in Tan A positionthe deflection produced is 600 . The displacement of the magnet to reduce the deflection by 300 isXH Ú_È"³á# x@x°=ôQê L#ß ÎO_È JÇ°ªøO`xß Tan A ªÖ#OÕ J=°~¡ó|_ÈÛ XH J¬=~¡ë# JÇ°ªøO Ç=¶¬HO g° Î

`Ƕ~¡°æ"³á¬ô J=°~¡ó|_#¬ôæ_È° L#ß J¬=~¡ë#O 600 J~ò Í P J¬=~¡ë#O 30

0 ÇQê¾O>è P PÇ°ªøO Ç=ò ³òHø

ªÖ#ãO×O

1) 17.32 cm 2) 10 cm 3) 7.32 cm 4) 34.64 cm

108. Two uncharged condensers A and B of capacities 3 µ F and 6 µ F are joined as shown. The free endsof the condensers are maintained at potentials of +12 V and -12 V. The potential at the junction O is

3 µ F =°iÇò 6 µ F ïH²\©° Q® A =°iÇò B J<Í ï~O_È° J<"Ít Ç HO_³#~¡°Á ¬@=òÕ KǶ²# q Î=òQê

He²<~¡°. P HO_³#~¡°Á z=~¡ =^Îí +12 V =°iÇò -12 V Ú>ÿxÂÇ°° LO_Í@@°ÁQê KͲ<~¡°. J~ò# O J<Í

aO ΰ=ô = Îí L#ß Ú>ÿxÂÇ° q°=

A BO

3 µF

−12 V

6 µF

+12 V

1) 4 V 2) - 4 V 3) zero 4) 8 V

109. A condenser of capacity C is charged to a potential difference V and the energy stored in thecondenser is U. A graph is drawn with V2 on X-axis and U on Y-axis, the graph is straight linethrough the origin. The slope of the straight line is

C ïH²\© Q® HO_³#~#° V J#° Ú>ÿxÂÇ° è^Î=ò#° P"Í׬iz<~¡°. P HO_³#~Õ xÞ L#ß ×H÷ë U.

X-JHÆO ¬~¡OQê V2 =°iÇò Y-JHÆO ¬~¡OQê U #° f¬°Hùx ãQꦬô wÇ°Qê Jk =üaO^ΰ=ô #°O_ "³×äKÇ°#ß

¬~¡×ö~Y ~¡¶¬=òÕ L#ßk. J~ò# P ¬~¡×ö~Y ³òHø "°

1) C 2) 2/C 3) C/2 4) 2C

(22)

110. An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistanceR that must be put in series with the bulb, so that the bulb draws 500 W is

500 W, 100 V |°Ä 200 V ¬ÌÁåä½ H¬|_#k. P |°Ä 500 W "³°=iOKÇ=ÿ#O>ÿ ^¥xH÷ ãõ÷Õ H¬=²#

x~Ë Î=ò

1) 188 2) 208 3) 408 4) 7008

111. The emf of a cell is 2V and its internal resistance is 2Ω . A resistance of 8Ω is joined to battery inparallel. This is contacted in secondary circuit of potentiometer. If 1 v standard cell balances for100 cm of potentiometer wire, the balance point of above cell is

2V q.K.| =°iÇò 2 Ω JO`Çiß~Ë^Î=ò Q® XH ¦¬°@=ò#ä½ 8 Ω x~Ë^Î=ò#° ¬=¶O`Ç~¡=òQê He² ^¥xx

Ú>ÿx³¶g°@~¡° ³òHø QÒ=Ç°=ò#ä½ He²<~¡°. 1 v ªO_È~Û ¦¬°@=ò#ä½ ¬O Ç°# Ú_È=ô 100 cm J~ò# Ìá

J=°iHä½ ¬O Ç°# Ú_È=ô

1) 120 m 2) 240 cm 3) 160 cm 4) 116 cm

112. The e.m.f. produced in a thermocouple is proportional to the temperature difference between thejunctions. When the temperature difference is 500C, thermo e.m.f. is 20mV. If the temperature ofthe cold junction is increased by 50C and that of hot junction is decreased by 50C then the changein thermo e.m.f. isL+¬âÇòQ®àOÕx q.K.| ¬Ok LéâãQ® Ç è Î=ò#ä½ J#°Õ=¶#° Ç=òÕ LO_È°#°. 500C LéâãQ®` è^Î=ò#ä½ L+¬â

q^ΰ¼`KH|=ò 20mV. KÇÁx ¬Ok LéâãQ®`Ç 50C ÌOz "Í_ ¬Ok LéâãQ®`Ç 50C Çy¾Oz# L+¬â q.K.| Õx

=¶~¡°æ

1) 20% decrease 20% ÇQ®°¾#° 2) 20% increase 20% Ì~¡°Q®°#°

3) 25% decrease 25% ÇQ®°¾#° 4) 25% increase 25% Ì~¡°Q®°#°

113. Arrange the devices, given below, in ascending order of their resistancesãH÷O Î Wzó# ¬iH~¡=ò ³òHø x~Ë Î=ò ¬~¡OQê P ¬iH~¡=ò#° P~ˬì ãH=°OÕ ã"Ç°O_.

a) Ammeter of range 1 mA 1 mA J=k Q® Jg°à@~

b) Ammeter of range 10 mA 10 mA J=k Q® Jg°à@~

c) Voltmeter of range 1 V 1 V J=k Q® Fg°@~

d) Voltmeter of range 10 V 10 V J=k Q® F g°@~

1) a, b, c, d 2) b, a, c, d 3) d, c, b, a 4) c, b, d, a

114. A proton with kinetic energy K describes a circle of radius r in a uniform magnetic field. An α -particle with kinetic energy K moving in the same magnetic field will describe a circle of radius¬=°su u~¡¼H± JÇ°ªøO Ç öHÆã Ç=òÕ K Q®u[×H÷ë He¾# ãé\ì#°#° ã¬H÷Ƭë=ò KͲ#¬ô_È° r J<Í "¼ª~¡=ò Q® =$`ëHê~¡

¬ ÎOÕ u~¡°Q®°KÇ°#ßk. J~ò# J Í JÇ°ªøO Ç öHÆã Ç=òÕ K J<Í Q®u[×H÷ë Q® α -H=ò#° ã¬H÷Ƭë=ò KͲ# ³°_È

P =$`ëHê~¡ ¬ Î=ò ³òHø "¼ª~¡=ò

1) 2

r2) r 3) 2r 4) 4r

115. The current and voltage in an a.c circuit containing inductance and resistance are given byi = 2 sin 314 t ampers and v = 60 sin (314t+ π /3) volts respectively. The value of the resistance in thecircuit is

ãÀ~¡H=ò =°iÇò x~Ë^Î=ò Q® AC =Ç°=ò #O^ΰ ã¬=²ìOKÇ°KÇ°#ß q^ΰ¼` =°iÇò ×Hà=ò q°=°

i = 2 sin (314 t) A =°iÇò v = 60 sin (314t+ π /3) V J~ò# P =Ç°=òÕ L#ß x~Ë Î=ò q°=

1) 15 Ω 2) 14 3Ω 3) 30 Ω 4) 20 Ω

(23)

116. X-rays of wavelength 0.1 A are allowed to fall on a metal scatterer. The wavelength of scattered

radiation is 0.111 A. The angle of scattering is

0.1 A0 Ç~¡OQ®^³á~¡É¼O Q® X-H÷~¡ì#° XH Õ¬ì ¬iöHƬH=ò g°^Î ¬`Ç#=ò K³Ok#¬ô_È° ¬iöHƬH qH÷~¡ Ç~¡OQ®^³á~¡É¼O

0.111 A0. J~ò# ¬iöHƬH HË=ò

1) 1 13

cos24

R

¬­ ­ ­­ ® 2) 1 11

cos24

R

¬­ ­ ­­ ® 3) 1 11

cos13

R

¬­ ­ ­­ ® 4) 1 11

cos20

R

¬­ ­ ­­ ®

117. Which of the following reactions is one of the reactions in carbon - nitrogen cycle for the emission ofenergy in sun and starsãH÷O^Î Wzó# KÇ~¡¼Õ XH KÇ~¡¼ Hê~¡Ä<£ Ð <³áã\Õ[<£ KÇãH=ò ^¥Þ~ ¬¶~¡°¼x #°O_ =°iÇò #HÆã`Ç=ò #°O_ "³°=_È°

×H÷ë ¬OOkOz# ¬g°H~¡=òJk U KÇ~¡¼

1) 1 2 31 1 2H H He+ → + energy 2) 1 15 12 4

1 7 6 2H N C He+ → +

3) 9 4 12 14 2 6 0Be He C n+ → + 4)

1 3 4 01 2 2 1H He He e+ → + + γ + energy

118. In a transistor the current amplification factor B’ ’ is 0.9. The transistor is connected in commonemitter configuration, the change in collector current when base current changes by 4 mA is

ã\ìx¬~ä½ L=°à_ P ¥~¡ q<¼¬=òÕ H~¡O@° =$k 0.9. D ã\ìx¬~¡° L=°à_ L^¥¾~¡ q<¼¬=òÕ L#߬C_È° P ¥~¡

H~¡O@°Õ =¶~¡°æ 4 mA J~ò# ÀHi÷ H~¡O@°Õ =¶~¡°æ.

1) 4 mA 2) 12 mA 3) 24 mA 4) 36 mA

119. The torque of a force ˆ ˆ ˆF 3i j 5k

G

acting at a point is UG . If the position vector of the point is

ˆ ˆ7i 3 j k , then UG is

XH aO^ΰ=ô ³òHø ªÖ# ¬k× ˆ ˆ7i 3 j k =°iÇò |=ò

ˆ ˆ ˆF 3i j 5k

G

J~ò# \ì~¡°ø

1) ˆ ˆ ˆ7i 8 j 9k 2) ˆ ˆ ˆ14i j 3k 3) ˆ ˆ ˆ2i 3j 8k 4) ˆ ˆ ˆ14i 38 j 16k

120. In Young's double slit experiment, the separation between slits is 2 x 103 m whereas the distance of

screen from the plane of slits is 2.5 m. Light of wavelengths in the range 2000 0A to 8000

0A is

allowed to fall on the slits. The wavelengths in the visible region that will be present on the screen at103m from central maximum is

Ç°OQ· kÞpeH 㬳¶Q®=ò#O^ΰ peH =°^μ L#ß ^ζ~¡O 2 x 103 m =°iÇò peH° =°iÇò ³~¡ä½ =° μ L#ß

ζ~¡O 2.5 m. J~ò Í 2000 0A #°O_ 8000

0A `Ç~¡OQ®^³á~¡É¼ "¼²ë Q® HêOu H÷~¡=ò° P peH g°^Î ¬_#¬C_È°

=°^μ ^ΰ¼u=°Ç° ¬\© #°O_ 103m ^ζ~¡OÕ ^Î$×¼ HêOuÕ U `Ç~¡OQ®^³á~¡É¼=ò° `³~¡g°^Î H#¬_È°`Ç°Ok.

1) 0

3500 A 2) 0

4000 A 3) 0

6000 A 4) 0

5000 A

(24)

CHEMISTRY

121. First ionisation energy of Li is 5.4eV and electron affinity of fluorine is 3.4eV. H∆ of the following

reaction, assuming that the ions are infinite distance apart is ( ) ( )( ) ( ) g gLi g F g Li F+ −+ → +

Li ã¬^Î=° PÇ°hH~¡ ×H÷ë 5.4eV =°iÇò ¦éÁi<£ Z¢Hê<£ Z¦²x\÷ 3.4eV°. JǶ#°° J#O Ç Î¶~¡ =¼`¼¬OÕ

L#ß=x ìqÀë, ( ) ( )( ) ( ) g gLi g F g Li F+ −+ → + KÇ~¡¼ä½ H∆(1) 23.06 K.Cal/mole 23.06 H÷.Hê/"³¶ (2) 46.12 K.Cal / mole 46.12 H÷.Hê/"³¶

(3) + 52 K.Cal/ mole + 52 H÷.Hê/"³¶ (4) 10 K.Cal/mole 10 H÷.Hê/"³¶

122. Correct statements in the following D ãH÷Ok "\÷Õ ¬i³Ø°# "¼Y¼°

a) 4NH Cl contain ionic, covalent and dative bonds

4NH Cl Õ JǶxH, ¬=°³¶[hÇ° =°iÇò _Í\÷"£ |O ¥°<ß~ò

b) As per valence bond theory 4CH molecule contain one s s− and three s p− "Íh |O^Î ²^¥O`Ç ã¬Hê~¡O 4CH J°=ôÕ XH s s− =°iÇò =ü_È° s p− |O ¥° L<ß~ò

c) Ethylene molecule contains 5 and one bonds Wkb<£ J°=ôÕ 5 =°iÇò XH |O Î=ò LO\ì~ò

d) As per VSEPR theory 3NH molecule is tetrahedral in shape VSEPR ²^¥O`ÇO ã¬Hê~¡O, 3NH J°=ô KÇ Ç°~¡°àvÇ° PHê~¡O Hey LOk

1) All Jhß 2) a Ê b only a Ê b =¶ã Ç"Í°

3) a Ê c only a Ê c =¶ã Ç"Í° 4) a, b and c only a, b =°iÇò c =¶ã Ç"Í°

123. The formal charge on 1st, 2nd nitrogen atoms and oxygen atom in N ON respectively

N ON Õ 1=, 2= <³áã\Õ[<£ =°iÇò PH÷[<£ ¬~¡=¶°=ôÌá ¦~¡à P"Í×O =~¡°¬Qê

1) -1, +1, 0 2) -1, 0, +1 3) +1, 0, -1 4) 0, -1, +1

124. Match the following [ Ǭ~¡°KÇ°=ò

List - I *ìa` Ð I List - II *ìa` Ð II

A) Blackash 1) used as water softner

#Áx |¶_ Î h\÷ Hêi#¼O `ùyOKÇ°@ä½ "_È°`~¡°

B) Soda lime 2) used as fire extinguisher

ªé_¨ÿá"£° =°O@#° P~¡°æ@ä½ "_È°`~¡°

C) Washing soda 3) used as decarboxylating agent

|@ ªé_¨ _Hê~ÄH÷bH~¡ Hê~¡HOQê "_È°`~¡°

D) Baking soda 4) a mixture of 2 3Na CO and CaS

èH÷OQ· ªé_¨ 2 3Na CO =°iÇò CaS q°ã×=°O

5) a mixture of 3CaCO and 2Na S

3CaCO =°iÇò 2Na S q°ã×=°O

The correct matching is ¬i³Ø°# ä~¡°æ

1) A - 4, B - 3, C - 1, D - 2 2) A - 5, B - 3, C - 2, D - 1

3) A - 4, B - 3, C - 2, D - 1 4) A - 5, B - 3, C - 1, D - 2

(25)

125. The angular momentum of an electron in hydrogen atom is 34 2 12.12 10 seckg m− −× then the electron

belongs to

Ìáìã_Ë[<£ ¬~¡=¶°=ôÕx XH Z¢Hê<£ HË©Ç° ã^Î=¼"ÍQ®=ò 342.12 10−× H÷ãQê.g°

2

.ÌÐ1

. J~ò# P Z¢Hê<£

______________ ä½ K³O^ΰ#°

1) M - shell M - HHÆ ¼ 2) K - shell K - HHÆ ¼

3) L - shell L - HHÆ ¼ 4) N - shell N - HHÆ ¼

126. 0.48 grams of hydrocarbon on combustion gave 0.66 grams of carbondioxide. The percentage ofcarbon and hydrogen respectively is

0.48 ãQê Ìáìã_ËHê~¡Ä<£#° άì#O K³OkOKÇQê 0.66 ãQê Hê~¡Ä<£_ÈǶïHå_£ U~¡æ_#k. J~ò# Hê~¡Ä<£ =°iÇò

Ìáìã_Ë[<£ ì~¡§`Ç=ò° =~¡°¬Qê

1) 62.5, 37.5 2) 37.5, 62.5 3) 45, 55 4) 55, 45

127. 4KMnOalcKOH3 3 alkaline21mole CCH CH A B C

A ¶¶¶¶l ¶¶¶l ¶¶¶l . Here 'C' is _____ WO Î°Õ 'C' J#°#k

1) Glycol ïQåÁHê 2) Carboxylic acid Hê~ÄH÷eH± P=°ÁO

3) Amine Zg°<££ 4) Aldehyde PeÌáì_£

128. 2 500water Fe X

CCaC A B Y°→ → → . The correct combination about X and Y in the following

X =°iÇò Y ^Î$¼ ¬i³Ø°# ä~¡°æ

1) If X is 2Cl in UV light then Y is 6 6 6C H Cl

X J#°#k UV HêOu ¬=°HÆOÕ 2Cl J~ò Í Y J#°#k 6 6 6C H Cl

2) If X is conc 3HNO + conc 2 4H SO at 40 C° then Y is m - dinitro benzene

X J#°#k 40 C° =^Îí Qê_È 3HNO Qê_È 2 4H SO J~ò Í Y J#°#k m - _³á <³áã\Õ ÿOr<£

3) If X is 3 2O H O+ then Y is adipic acid X J#°#k 3 2O H O+ J~ò# Y J#°#k Z_²H± P=°ÁO

4) If X is 3 3CH COCl AlCl+ then Y is toluene X J#°#k 3 3CH COCl AlCl+ J~ò Í Y J#°#k \Õb<£

129. Equivalent weight of 3 4H PO in a reaction of 3 4 2 4 22 2NaOH H PO Na HPO H O+ → + is

(M is molecular weight of 3 4H PO )

3 4 2 4 22 2NaOH H PO Na HPO H O+ → + J#° K Ç~ ¡ ¼Õ 3 4H PO ³òH ø ` Ç °¼ ì~ ¡=ò

( 3 4H PO J°ì~¡=ò M)

1) 3

M2) M 3)

2

M4)

4

M

130. In the reaction, 2 32 7 2 3 2Cr O NO H Cr NO H O− − + + −+ + → + + the stoichiometric coefficients of

22 7 2,Cr O NO and H− − + are respectively.

2 32 7 2 3 2Cr O NO H Cr NO H O− − + + −+ + → + + J#° KÇ~¡¼ Ç°¼ ¬g°H~¡OÕ

22 7 2, ,Cr O NO H− − +

Q®°Hê° =~¡°¬Qê

1) 1, 3, 8 2) 1, 4, 8 3) 2, 3, 8 4) 1, 5, 12

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131. The ratio of average kinetic energy 2O at 0 C° to 2SO at 273 C° is

0 C° =^Îí Q® 2O =°iÇò 273 C° =^Îí Q® 2SO ¬Q®@° Q®u[×ä½ë x+¬æuë

1) 2 : 1 2) 1 : 2 3) 1 : 4 4) 4 : 1

132. Incorrect statements are D ãH÷Ok "\÷Õ ¬iHêx "¼Y¼°

(A) 2 22BaCl H O is an interstitial hydrate 2 22BaCl H O J#°#k JìæO Ç~× Ìáìã_Í\

(B) 4 2 4 25 5CuSO H O CuSO H O+ → is an example of hydrolysis

4 2 4 25 5CuSO H O CuSO H O+ → J#° KÇ~¡¼ [qõÁ+¬ìxH÷ L^¥¬ì~¡

(C) Iron-oxide is used to remove the sulphur during 'Fischer-Tropsch' process

¦²+¬~ - \ì~æ¹ q^¥#OÕ ¬æ~ =°e<xß `ùeyOKÇ°@ä½ S~¡<£ PïHá _£#° L¬³¶yªë~¡°

(D) 3 2 3 33 3PCl H O H PO HCl+ → + is an example of hydration

3 2 3 33 3PCl H O H PO HCl+ → + J#° KÇ~¡¼ ªãsÛH~¡=ò#ä½ L^¥¬ì~¡

1) A,B Ê C only A,B Ê C =¶ã Ç"Í° 2) A,B Ê D only A,B Ê D =¶ã Ç"Í°

3) B Ê D only B Ê D =¶ã Ç"Í° 4) A Ê C only A Ê C =¶ã Ç"Í°

133. Correct statements regarding silicones are ²eHË<£ q+¬Ç°Õ ¬i³Ø°# "¼Y¼ (°)

1.They are used in the preparation of water proof clothes g\÷x "@~ 㦬 |@ ÇǶsÕ "_È°`~¡°

2.They are organosilicon compounds Wq P~¾<˲eHê<£ ¬"Í°à×<°

3. They are used in the preparation of grease and lubricants g\÷x HO ³#° Ê ãwA ÇǶsÕ "_È ~¡°

4. All Jxခ

134. Correct statements among the following D ãH÷Ok "\÷Õ ¬i³Ø°# "¼Y¼°

a) 95 - 98% pure boron is called Moissan Boron 95 - 98% ×à ΠÕ~<£ #° =¶~òª<£ Õ~<£ JO\ì~¡°

b) Pure Boron is used as semi conductor ×Ã^Î Õ~<£#° J~¡Ö "¬ìHOQê "_È`~¡°

c) Oxides of boron are amphoteric in nature Õ~<£ PïHá _£° kÞ¬Þì=O He¾ LO\ì~ò

1) a and b are correct a =°iÇò b ¬i³Ø°#q 2) b and c are correct b =°iÇò c ¬i³Ø°#q

3) a and c are correct a =°iÇò c ¬i³Ø°#q 4) all are correct Ìá=xßÇü

135. Regarding 2XeF , the incorrect statement 2XeF Q®°iOz ¬iHêx "¼Y¼

1. Hybridisation is 3sp d 3sp d ¬OHsH~¡ Hk

2. Shape is linear ö~vÇ° PH$u Hk

3. There are two lone pairs on central atom =°^μ¬Ö ¬~¡=¶°=ô Ìá ï~O_È° XO@i [O@°<ß~ò

4. It is formed by 'Xe' in first excited state nxx Xe' ã¬^Î=° L`Íël`Ç ²ÖuÕ U~¡æ~¡°¬°ëOk

136. In activated carbon method for defluoridation the exhausted filters are reactivated by washing firstwith ----------- and then with --------------

_¦éÁi_Í+¬<£ ä½ L¬³¶yOKÇ° L Íël Ç Hê~¡Ä<£ q ¥#OÕ qx=°Ç° ¦²~#° "³ò Î@ ..................... ¥Þ~ Ç Î°¬i

..................... ^¥Þ~ H_ÈQ®_ÈO =Á ¬ô#~¡° Îiªë~¡°

1) 4% NaOH, 1% H3PO4 2) 1% H3PO4, 4% NaOH

3) 1% HCl, 4% NaOH 4) None of these UnHê^ΰ

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137. Assertion (A) : The electronic configuration of Chromium is ( ) 5 13 4Ar d s but not ( ) 4 23 4Ar d s

xtó`Ç "¼Y¼ (A) : ãHËq°Ç°O ZãHêxH± q<¼¬O( ) 5 13 4Ar d s Hêh ( ) 4 23 4Ar d s Hê ΰ

Reason (R) : The lowering of energy due to exchange pairs of electrons is more in ( ) 4 23 4Ar d s

than in ( ) 5 13 4Ar d s

q=~¡ (R) : ZãHê<£ [O@ ¬~¡¬æ~¡ =¶iæ_ =Á ( ) 4 23 4Ar d s Õx ×H÷ë `ÇQ®°¾^Î( ) 5 13 4Ar d s Õ H<ß Zä½ø=

1) Both A and R are true and R is the correct explanation of A.

A =°iÇò R ° ï~O_ȶ x["³°Ø#q. R J#°#k A ä½ ¬i³Ø°# q=~¡.

2) Both A and R are true and R is not the correct explanation of A.

A =°iÇò R ° ï~O_ȶ x["³°Ø#q. R J#°#k A ä½ ¬i³Ø°# q=~¡ Hê^ΰ.

3) A is true, R is false A X¬ôC, R ǬôC 4) A is false but R is true. A ǬôC,R X¬ôC

138. Iron reacts with 4CuSO according to the equation 4 4Fe CuSO FeSO Cu+ → + . If excess of

iron is added to 500ml of 0.2M 4CuSO solution, the amount of copper formed is ______ (At.wt of

Cu is 64)

4CuSO Ë S~¡<£, 4 4Fe CuSO FeSO Cu+ → + J#° ¬g°H~¡O ã¬Hê~¡O KÇ~¡¼ [i¬ô Ç°Ok. 500ml 0.2M

4CuSO ã^¥=ìxH÷ Zä½ø= S~¡<£#° He² Í, U~¡æ_# Hê¬~ ì~¡=ò (Cu ¬.ì = 64)

1) 6.4 gm 2) 3.2 gm 3) 0.64 gm 4) 0.32 gm

139. 520K 870KX Y Z¶¶¶l ¶¶¶l . 'Y' is tetrabasic acid but gives only two series of salts, then X, Y & Z arerespectively

520K 870KX Y Z¶¶¶l ¶¶¶l 'Y' J#°#k ï~O_È° = ãõ°#° =¶ã Ç"Í° U~¡æ~¡°KÇ° KÇ Ç°HÆê~¡ Ç P=°Á=ò. J~ò# X, Y

& Z ° =~¡°¬Qê

1) H3 PO4 , HPO3 , H4P2O7 2) HPO3 , H3PO4 , H4P2O7

3) H3 PO4 , H4P2O7 , HPO3 4) H3 PO4 , H4P2O6 , HPO3

140. The work done on the surroundings is 9 joules. 45J heat is supplied to the system. The change ininternal energy is1) –36J 2) –54J 3) +36J 4) +54 J

XH =¼=¬Öä½ 45J Lâxß JOkOz#¬ô_È° Jk ¬i¬~Ìá 9 *º° ¬xx [i²#>ÿåÁ`Í =¼=¬Ö Õ Hey# JO`ÇiH

×H÷ë =¶~¡°æ

1) –36J 2) –54J 3) +36J 4) +54 J

141. Average molecular weight ( )wM of a polymer can be determined experimentally by

XH e=°~ ³òHø ¬Q®@° ì~¡ J°ì~¡O ( )wM #° 㳶yHOQê D q ¥#OÕ x~Öiªë~¡°

1) Victor - Mayer method qH~ Ð "Í°Ç°~ ¬^Îu 2) viscosity measurements ²ßQ® Hù Ç°

3) scattering of light method HêOu qöHÆƬ q ¥#O 4) Both (2) and (3) (2) =°iÇò (3)

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142. ( ) ( ) 32 23 2 92g gN H NH H KJ+ → ∆ = − . Then the heat of formation of ammonia is

( ) ( ) 32 23 2 92g gN H NH H KJ+ → ∆ = − J#° KÇ~¡¼ ^¥Þ~ J"³¶àxǶ ³òHø ¬OõÁé+¬â=ò

1) - 92 KJ 2) - 46 KJ 3) + 92 KJ 4) + 46 KJ

143. An inter halogen compound is formed by bromine in it's second excited state. The shape of thecompound is

XH JO`Ç~¡ ¬ðÕ[<£ ¬"Í°à×#=ò#° ãÕq°<£ ^¥x ï~O_È= L`Íël`Ç ²ÖuÕ U~¡æiÀë, P ¬"Í°à×#=ò PH$u

1) T - shape T - PHê~¡=ò 2) square pyramidal KÇ Ç°~¡ã¬ ²~¡q°_È

3) pentagonal bi-pyramidal ÌO\ìQË# kÞ²~¡q°_È 4) square bipyromidal KÇ Ç°~¡ã¡ kÞ²~¡q°_È

144. For an electrolyte solution of 0.05 mol L-1, the conductivity has been found to be 0.0110 S cm-1. Themolar conductivity is

0.5 "³¶.b-1 Qê_È`Ç Q® q^ΰ¼kÞõÁ+¬¼ ã^¥==ò ³òHø "¬ìH`Ç 0.0110 S cm-1

. J~ò ÍÇ ¥x "³¶ì~ "¬ìH Ç q°=

1) 0.055 S cm2mol-1 2) 550 S cm2 mol-1 3) 0.225 cm2 mol-1 4) 220 S cm2mol-1

145. Which of the following carbohydrate has 1 2→ glycosydic bond

1 2→ ïQåÁHËÌá_H± |O^Î=ò#° Hey L#ß Hê~ËÄÌáìã_Í\ ãH÷Ok "\÷Õ

1) Maltose =¶Õ* 2) Sucrose ¬°ãHË*

3) Lactose ìHË* 4) Cellulose Ì°¼Õ*

146. The structure of Thymine is ãH÷Ok "\÷Õ ³ág°<£ x~à=ò

1) NH

NH

O

O

2)

3CHNH

NH

O

O

3) N

NH

NH2

O

4) NH

NNH

N

O

NH2

147. For flocculation of As2S

3 sol, among the following electrolytes which one is most effective

As2S

3 ª#° ¬øO Î#O K³OkOKÇ°@ä½, D ãH÷Ok "\÷Õ JkH ª=°~¡¼=ò Q® q ΰ¼ qõÁ+¬¼O

1) 4 6K Fe CN ¯

¡ °¢ ± 2) 3 4Na PO 3) 2 4 3A SOA 4) MgCl2

148. Ammonium carbamate is allowed to decompose in a closed container. If equilibrium pressure for

4 2 3 2S g gNH COONH 2NH COU at certain temperature is 0.9 atm. Then, partial pressureof Ammonia at equilibrium (in atm)

XH =ü² LOz# ã ÇÕ J"³¶àxÇ°O Hê~ùÄ"Í°\ q¦¬°@#=ò K³OkOKÇ#¬ô_È°,

4 2 3 2S g gNH COONH 2NH COU J#° ¬=°`²Öu Hey#k. ¬=°`²Öu Ô_È#O 0.9 J\ìà J~ò#

¬=°`²Öu = Îí J"³¶àxǶ H÷ÆH Ô_È#O (J\ìàÕ)

1) 0.9 2) 0.81 3) 0.03 4) 0.6

149. The effective number of atoms present in the unit cell of a simple cube, body centered and facecentered cubic cell are respectively

ª^¥~¡ ¦¬°#=ò, JO`ÇöHOãk`Ç ¦¬°#=ò, ¦¬H öHOãk`Ç ¦¬°< Çüx\ Ì Õx ãìqHÇ Hì ¬OY¼ =~¡°¬Qê

1) 1, 1, 1 2) 1, 2, 2 3) 1, 2, 3 4) 1, 2, 4

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150. A : Argentite is leached with the solution of NaCN in presence of air

xtó Ç "¼Y¼ (A) : Qêe ¬=°HÆOÕ Pï~ûO>ÿØ\#° NaCN ã^¥=O`Ë bzOQ· Kͪë~¡°

R : O2 of air oxidises Na

2S to Na

2SO

4 and prevents the reverse reaction

- q=~¡ (R) : QêeÕx O2, Na2S #° Na2SO4 Qê PH©HiOz u~ËQêq° KÇ~¡¼#° x~Ëk¬°ëOk

1) Both A and R are true and R is the correct explanation of A. A =°iÇò R ° ï~O_ȶ x["³°Ø#q. R J#°#k A ä½ ¬i³Ø°# q=~¡.

2) Both A and R are true and R is not the correct explanation of A.

A =°iÇò R ° ï~O_ȶ x["³°Ø#q. R J#°#k A ä½ ¬i³Ø°# q=~¡ Hê^ΰ.

3) A is true, R is false A X¬ôC, R ǬôC 4) A is false but R is true. A ǬôC,R X¬ôC

151. The complexes [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br can be identified by

[Co(NH3)5Br]SO4 =°iÇò Co(NH3)5SO4]Br J<Í ¬OtÁ#° Q®°iëOKÇ°@Õ "_È|_Íq

I) AgNO3 II) CuSO4 III) BaCl2Correct answer is ¬i³Ø°# ["|°

1) I only I =¶ã Ç"Í° 2) II only II =¶ã Ç"Í°

3) I and III only I =°iÇò III =¶ã Ç"Í° 4) I and II only I =°iÇò II =¶ã Ç"Í°

152. 0.004M solution of ( )4 6K Fe CN is isotonic with 0.014M solution of urea at same temperature.

The degree of dissociation of ( )4 6K Fe CN is

0.004M ( )4 6K Fe CN ã^¥=O =°iÇò 0.014M ÇüiǶ ã^¥=ì° Sªé\ÕxH±Qê L#æ¬C_È°

( )4 6K Fe CN ³òHø q¦¬°@# fã= Ç

1) 50% 2) 62.5% 3) 75% 4) 90%

153. The following graph is true for a second order reaction

ï~O_È= ãH=¶OH KÇ~¡¼ä½ ¬i³Ø°# ãQꦹ ãH÷Ok "xÕ

1) 2)

3) 4)

154. A buffer solution containing 50 ml 0.1M 3CH COOH and 50ml 0.1M 3CH COONa is diluted

by adding 900ml water. The pH of resulting solution is ___________ (For

3 4.8KaCH COOH P = )

50 q°.b. 0.1M 3CH COOH =°iÇò 50 q°.b 0.1M 3CH COONa Q® |¦¬~ ã^¥==ò#°

900q°.b. h\÷ Ë qb#O KÍÇ°Qê U~¡æ_# ¦¬e Ç ã^¥==ò pH q°= ( 3CH COOH ä½ 4.8KaP = )

1) 3 2) 4.8 3) 7 4) 6.8

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155. If the molar solubility of a sparingly soluble salt 3MX is 'a' mole lit then solubility product of

3MX is XH Jæã^¥=©Ç°`Ç Q® 3MX=O ³òHø ã^¥=©Ç° Ç a "³¶ b

Ð1

J~ò Í 3MX ³òHø ã^¥=©Ç° Ç |=ò q°=

1) 43spK a= 2) 427spK a= 3) 327spK a= 4) 33spK a=

156.2 5

2 5 3

C H ONaAlc KOH HClAnhydrous AlCl

C H Cl X Y Z−∆− → → →

)K=ä¢×*

. Compound 'Z' is

¬"Í°à×#O 'Z' J#°#k

1) dimethyl ether _³á q°^³á D^Î~ 2) Ethyl acetate W^³á Z²>è\

3) diethyl ether _³áW^³á D^Î~ 4) Methyl formate q°^³á ¦ö~à\

157. In Victor Meyer's test, alcohol is mixed with red 2 2&P I AgNO+ and nitrous acid, the resultant'X' is mixed with alkali to give red colour. Then compound 'X' isqH~ "Í°Ç°~ ¬^ÎuÕ Pø¬ð#° Zã~¡ 2 2&P I AgNO+ =°iÇò <³áã@¹ P=°ÁO`Ë KÇ~¡¼ [i²Oz U~¡æ_# 'X' #°

HÆê~xH÷ He²`Í Zã~¡x ~¡OQ®° U~¡æ_Ok. J¬ôæ_È° ¬"Í°à×#O 'X' J#°#k

1) 2 2R CH NO− − 2) 2R C NO− −

N OH−3)

2 2R C NO− −

NO4)

R C OH− −

2N NO−

158. The correct sequence of reactions involved in the nitration of aniline are respectivelyZxe<£ #° <³áã\ÕH~¡O K³OkOKÇ_ÈOÕ [~¡°Q®° KÇ~¡¼ ¬~¡O¬~¡ ãH=°=ò

1) Acylation, nitration, hydrolysis ZÌáè+¬<£, <³áã\ÕH~¡O [qõÁ+¬

2) Hydrolysis, acylation, nitration [eqõÁ+¬, [qõÁ+¬, <³áã\ÕH~¡O

3) Acylation, hydrolysis, nitration ZÌáè+¬<£ , [qõÁ+¬ , <³áã\ÕH~¡O

4) Alkylation, nitration, hydrolysis Pÿáøb H~¡O, <³áã\ÕH~¡O , [qõÁ+¬

159. Compound A (C5H

10O) forms a phenyl hydrazone and gives negative response for Tollens test and

iodoform test. Then the compound A is¬"Í°à×#=ò A (C5H10O) J<Ík ¦²<³á Ìáìã_È*Õ<£#° U~¡æ~¡°¬°ëOk Hêh \çÿ<£ ¬sHÆ =°iÇò J³ò_˦O ¬sHÆÕ

ç¾# ΰ. J~ò Í A J#° ¬"Í°à×#O

1) 3 2 2 3

|

OH

CH CH CH CH CH− − − − 2) 3 2 2 3

||

O

CH CH CH C CH− − − −

3) 3 2 2 3

||

O

CH CH C CH CH− − − − 4) 2 2 2 2CH CH CH CH CH OH= − − −

160. Which of the following is used as "Morning after pill"ãH÷Ok "\÷Õ Íxx " =¶ißOQ· P¦¬~ ²" (Morning after pill) Qê "_³^Î~¡°

1) Norethindrone <iëOã_Ë<£ 2) Mife pristone "³°Ø㦲ªé<£

3) Novestrol <"³ãª 4) Lansoprazole ìOªã¬*Õ